1. Time Period From Frequency
Question: A body completes 5 oscillations in one second. Find its time period.
Show Solution
Given: f = 5 Hz.
Formula: T = 1/f.
Solution: T = 1/5 = 0.20 s.
Final Answer: 0.20 s.
If Periodic Motion or SHM is not clear and you are looking for a Physics Tutor, contact Kumar Sir.
Phone: +91-9958461445
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CLASS 11 PHYSICS
Periodic Motion and Simple Harmonic Motion
Class 11 Physics notes covering periodic motion, oscillatory motion, SHM, restoring force, graphical understanding, numericals and PYQs.
CBSENEETJEE MainJEE AdvancedIBICSEIGCSEA-Level
Introduction: Why Periodic Motion and SHM Matter
Periodic motion and simple harmonic motion are the grammar of repeated motion. Once a student understands this chapter, waves, sound, alternating current, resonance, molecular vibrations and many modern physics ideas become far more natural.
Teacher's View
In coaching classrooms, this chapter is not treated as a collection of formulas. It is taught as a way of seeing motion: a body moves away from equilibrium, a restoring influence pulls it back, inertia carries it to the other side, and the cycle repeats.
Real-life example: when a child on a swing is pulled back and released, the swing repeatedly crosses its middle position. That motion immediately introduces mean position, amplitude, time period, restoring tendency and energy exchange.
Exam Perspective
CBSE commonly asks definitions, graph interpretation and spring or pendulum applications. NEET prefers direct formula use plus concept traps. JEE Main asks fast mixed questions on equations, phase and energy. JEE Advanced may combine SHM with circular motion, constraints, springs and energy methods.
Memory trick: if the force says "come back" and becomes stronger when displacement becomes larger, start thinking about SHM.
Core idea: every SHM is periodic and oscillatory, but every periodic or oscillatory motion is not necessarily SHM.
Periodic Motion
Periodic motion is motion that repeats itself after equal intervals of time. The repeated time interval is called the time period.
Definition
A motion is called periodic if the state of the object repeats after a fixed time interval. The position, velocity and overall condition of the object return in the same pattern again and again.
Physical Meaning
Periodic motion means nature has a clock inside the motion. We do not need to watch the entire motion forever; after one full cycle, the same pattern restarts. The Earth around the Sun, the hands of a clock and the rotation of a fan all show repeated patterns.
Mathematical Interpretation
If a quantity x depends on time t and repeats after time T, then x(t + T) = x(t). The least positive value of T is the time period. Frequency is the number of cycles completed per second.
T = 1/f
T is time period in seconds. f is frequency in hertz. This formula says that if cycles are more frequent, each cycle takes less time.
Real-Life Examples
- Earth around Sun: nearly periodic with a time period of one year.
- Clock pendulum: repeats its motion after a fixed interval.
- Hands of clock: second hand repeats every 60 seconds, minute hand every hour.
- Rotating fan: each blade reaches the same orientation again after a fixed time.
- Planetary motion: planets return to similar positions in their orbits after fixed orbital periods.
Common Mistakes
- Thinking every repeated-looking motion is exactly periodic. Real systems may be approximately periodic due to friction.
- Confusing time period with frequency. Time period is time per cycle; frequency is cycles per second.
- Forgetting that uniform circular motion is periodic even though it is not to-and-fro motion.
f = 1/T
f means number of complete oscillations or cycles per second. Unit: hertz.
ω = 2πf
ω is angular frequency in rad/s. It connects one full cycle with 2π radians.
ω = 2π/T
This version is useful when the time period is directly given.
Oscillatory Motion
Oscillatory motion is to-and-fro motion about a mean position. The object crosses the mean position repeatedly and moves on both sides of it.
Definition and Mean Position
Oscillatory motion occurs when a body moves repeatedly on either side of a fixed position called the mean position or equilibrium position. The mean position is where the net restoring influence is zero.
Physical meaning: the body does not keep moving in one direction. It comes back, crosses the central point and goes to the other side. This repeated to-and-fro motion is the visible signature of oscillation.
Examples
- Swing: moves on both sides of its lowest position.
- Pendulum: bob moves left and right about the vertical line.
- Spring block: block moves left and right about natural length position.
- Tuning fork: prongs vibrate rapidly about their mean positions.
Oscillatory Motion About Mean Position
Exam Traps
An oscillatory motion may not be simple harmonic. For SHM, the restoring force must be proportional to displacement and opposite in direction. A pendulum with small angular displacement is treated as SHM, but a large-angle pendulum is not exactly SHM.
Simple Harmonic Motion Definition
Simple harmonic motion is a special oscillatory motion in which acceleration is directly proportional to displacement from the mean position and always directed towards the mean position.
Mathematical Condition
a = −ω²x
a is acceleration, ω is angular frequency, and x is displacement from mean position. The negative sign shows acceleration is opposite to displacement.
Restoring Force Form
F = −kx
F is restoring force, k is force constant, and x is displacement. The more the body is displaced, the stronger the restoring force becomes.
Physical meaning of the negative sign: if displacement is to the right, restoring force and acceleration act to the left. If displacement is to the left, restoring force and acceleration act to the right. The negative sign is not saying the force is small; it is saying the direction is opposite.
Hooke's Law Connection
A spring obeying Hooke's law gives F = −kx. With Newton's second law, ma = −kx, so a = −(k/m)x. Comparing with a = −ω²x gives ω² = k/m. This is why a spring-mass system is the cleanest model of SHM.
Examples
- A horizontal spring block on a smooth table.
- A simple pendulum for small angular displacement.
- A liquid column oscillating in a U-tube for small displacement.
- Small vibrations of atoms about equilibrium in a solid.
Common Mistake
Students often write F = kx and forget direction. In SHM, direction is everything. The correct restoring force form is F = −kx when x is measured from equilibrium.
Characteristics of SHM
Periodicity
SHM repeats after a fixed time period. A spring block returns to the same position and velocity after every complete oscillation.
Trap: position alone is not enough; velocity direction must also repeat for the full state to repeat.
Mean Position
The mean position is the equilibrium point. Net restoring force is zero here, but speed is maximum.
Example: the lowest point of a small-angle pendulum is its mean position.
Amplitude
Amplitude is maximum displacement from mean position. In ideal SHM, amplitude remains constant.
Memory trick: amplitude is the "edge distance" from center, not total distance between extremes.
Frequency
Frequency tells how many oscillations occur per second. A tuning fork of 256 Hz completes 256 vibrations per second.
Exam tip: convert milliseconds to seconds before using formulas.
Phase
Phase tells the stage of oscillation. Two bodies may have the same amplitude and frequency but different phases.
Example: two swings moving together are in phase; opposite motion gives phase difference π.
Energy Behaviour
Kinetic energy is maximum at mean position. Potential energy is maximum at extremes. Total energy remains constant in ideal SHM.
Trap: acceleration is maximum at extremes, not at mean position.
Restoring Force
Restoring force is the force that tries to bring a displaced body back towards its mean position.
Why It Acts Towards Mean Position
Mean position is stable equilibrium. When the body is shifted away, the system develops a force opposite to the shift. This force does not simply stop the body; it accelerates it back toward equilibrium. Due to inertia, the body crosses the mean position and reaches the other side.
Examples
- Spring: if stretched right, spring pulls left; if compressed left, spring pushes right.
- Pendulum: component of weight along the arc pulls the bob toward the lowest point.
- Liquid column: height difference creates pressure force that drives liquid back toward equal levels.
Restoring Force Direction in Spring-Mass System
Conditions for SHM
1. Stable Equilibrium
The object must have a mean position where it can remain at rest if undisturbed. Example: a pendulum at its lowest point. Non-example: a ball balanced at the top of a dome; small displacement makes it move away.
2. Restoring Force
After displacement, a force must arise toward the mean position. Example: spring force. Non-example: a stone moving freely in space after being displaced has no restoring force.
3. Proportional to Displacement
For exact SHM, the magnitude of force must be proportional to displacement. Doubling displacement doubles restoring force. This is why Hooke's law springs are ideal SHM systems.
4. Opposite Direction
Force must be opposite to displacement, written as F = −kx. A force in the same direction as displacement would push the object away and cannot produce SHM.
| System | Is it SHM? | Reason | Exam Trap |
|---|---|---|---|
| Ideal spring block | Yes | F = −kx exactly for an ideal spring. | Frictionless surface assumed unless stated. |
| Small-angle pendulum | Approximately yes | sin θ ≈ θ for small angles. | Large angle pendulum is not exact SHM. |
| Uniform circular motion | Projection is SHM | Projection satisfies x = A cos(ωt). | The circular motion itself is periodic, not to-and-fro SHM. |
| Bouncing ball | No | Force is not proportional to displacement throughout motion. | It repeats approximately but is not SHM. |
Graphical Understanding of SHM
Graphs are the fastest way to understand phase, maximum values and signs in SHM. Use HTML-safe formulas: x = A sin(ωt), x = A cos(ωt), v = Aω cos(ωt), a = −ω²x.
Displacement vs Time: x = A sin(ωt)
Velocity vs Time: v = Aω cos(ωt)
Acceleration vs Time: a = −ω²x
Force vs Displacement: F = −kx
Potential Energy vs Displacement: U = 1/2 kx²
Kinetic Energy vs Displacement: K = 1/2 k(A² − x²)
Formula Cards
x = A sin(ωt)
x: displacement, A: amplitude, ω: angular frequency, t: time. Starts from mean position.
x = A cos(ωt)
Useful when the body starts from an extreme position at t = 0.
v = Aω cos(ωt)
Velocity is maximum at mean position and zero at extremes.
a = −ω²x
Acceleration is opposite to displacement and proportional to displacement.
F = −kx
Restoring force in an ideal spring system. k is spring constant.
T = 2π/ω
Time period is time for one full oscillation.
Need Help With Periodic Motion or SHM?
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40 Solved Numericals
Solutions are hidden so students can attempt first. Open each solution only after writing the given data and selecting the correct formula.
2. Frequency From Time Period
Question: A pendulum has time period 2 s. Find frequency.
Show Solution
Given: T = 2 s.
Formula: f = 1/T.
Solution: f = 1/2 = 0.5 Hz.
Final Answer: 0.5 Hz.
3. Angular Frequency
Question: Find angular frequency if frequency is 4 Hz.
Show Solution
Given: f = 4 Hz.
Formula: ω = 2πf.
Solution: ω = 2π × 4 = 8π rad/s.
Final Answer: 8π rad/s.
4. Period From Angular Frequency
Question: If ω = 10π rad/s, find T.
Show Solution
Given: ω = 10π rad/s.
Formula: T = 2π/ω.
Solution: T = 2π/(10π) = 0.2 s.
Final Answer: 0.2 s.
5. Spring Angular Frequency
Question: A 0.5 kg block is attached to a spring of k = 200 N/m. Find ω.
Show Solution
Given: m = 0.5 kg, k = 200 N/m.
Formula: ω = √(k/m).
Solution: ω = √(200/0.5) = √400 = 20 rad/s.
Final Answer: 20 rad/s.
6. Spring Time Period
Question: For k = 100 N/m and m = 1 kg, find T.
Show Solution
Given: k = 100 N/m, m = 1 kg.
Formula: T = 2π√(m/k).
Solution: T = 2π√(1/100) = 2π/10 = π/5 s.
Final Answer: π/5 s.
7. Restoring Force
Question: A spring has k = 50 N/m. Find restoring force at x = 0.10 m.
Show Solution
Given: k = 50 N/m, x = 0.10 m.
Formula: F = −kx.
Solution: F = −50 × 0.10 = −5 N.
Final Answer: 5 N towards mean position.
8. Acceleration From Displacement
Question: In SHM, ω = 6 rad/s and x = 0.05 m. Find acceleration.
Show Solution
Given: ω = 6 rad/s, x = 0.05 m.
Formula: a = −ω²x.
Solution: a = −36 × 0.05 = −1.8 m/s².
Final Answer: 1.8 m/s² towards mean position.
9. Maximum Speed
Question: A particle has A = 0.20 m and ω = 10 rad/s. Find maximum speed.
Show Solution
Given: A = 0.20 m, ω = 10 rad/s.
Formula: vmax = Aω.
Solution: vmax = 0.20 × 10 = 2 m/s.
Final Answer: 2 m/s.
10. Maximum Acceleration
Question: A = 0.10 m and ω = 20 rad/s. Find maximum acceleration.
Show Solution
Given: A = 0.10 m, ω = 20 rad/s.
Formula: amax = ω²A.
Solution: amax = 400 × 0.10 = 40 m/s².
Final Answer: 40 m/s².
11. Speed at Displacement
Question: A = 10 cm, x = 6 cm, ω = 5 rad/s. Find speed.
Show Solution
Given: A = 0.10 m, x = 0.06 m, ω = 5 rad/s.
Formula: v = ω√(A² − x²).
Solution: v = 5√(0.01 − 0.0036) = 5 × 0.08 = 0.40 m/s.
Final Answer: 0.40 m/s.
12. Energy in Spring SHM
Question: k = 200 N/m, A = 0.10 m. Find total energy.
Show Solution
Given: k = 200 N/m, A = 0.10 m.
Formula: E = 1/2 kA².
Solution: E = 1/2 × 200 × 0.01 = 1 J.
Final Answer: 1 J.
13. Potential Energy at x
Question: k = 100 N/m and x = 0.04 m. Find potential energy.
Show Solution
Given: k = 100 N/m, x = 0.04 m.
Formula: U = 1/2 kx².
Solution: U = 1/2 × 100 × 0.0016 = 0.08 J.
Final Answer: 0.08 J.
14. Kinetic Energy at x
Question: k = 100 N/m, A = 0.10 m, x = 0.06 m. Find kinetic energy.
Show Solution
Given: k = 100 N/m, A = 0.10 m, x = 0.06 m.
Formula: K = 1/2 k(A² − x²).
Solution: K = 50(0.01 − 0.0036) = 0.32 J.
Final Answer: 0.32 J.
15. Mass From Period
Question: A spring of k = 400 N/m has time period π/10 s. Find mass.
Show Solution
Given: k = 400 N/m, T = π/10 s.
Formula: T = 2π√(m/k).
Solution: π/10 = 2π√(m/400), so 1/20 = √(m/400). Squaring gives 1/400 = m/400, hence m = 1 kg.
Final Answer: 1 kg.
16. Spring Constant From Period
Question: A 2 kg mass oscillates with T = π s. Find k.
Show Solution
Given: m = 2 kg, T = π s.
Formula: T = 2π√(m/k).
Solution: π = 2π√(2/k), so 1/2 = √(2/k). Squaring gives 1/4 = 2/k, so k = 8 N/m.
Final Answer: 8 N/m.
17. Pendulum Period
Question: Find time period of a simple pendulum of length 1 m. Take g = π² m/s².
Show Solution
Given: L = 1 m, g = π² m/s².
Formula: T = 2π√(L/g).
Solution: T = 2π√(1/π²) = 2 s.
Final Answer: 2 s.
18. Pendulum Length
Question: A seconds pendulum has T = 2 s. Find L if g = π² m/s².
Show Solution
Given: T = 2 s, g = π² m/s².
Formula: T = 2π√(L/g).
Solution: 2 = 2π√(L/π²), so 1 = π√(L/π²), hence L = 1 m.
Final Answer: 1 m.
19. Displacement Equation
Question: For x = 0.1 sin(20t), identify A and ω.
Show Solution
Given: x = 0.1 sin(20t).
Formula: Compare with x = A sin(ωt).
Solution: A = 0.1 m and ω = 20 rad/s.
Final Answer: A = 0.1 m, ω = 20 rad/s.
20. Find Period From Equation
Question: x = 5 sin(4πt) cm. Find T and f.
Show Solution
Given: ω = 4π rad/s.
Formula: T = 2π/ω and f = 1/T.
Solution: T = 2π/(4π) = 0.5 s; f = 2 Hz.
Final Answer: T = 0.5 s, f = 2 Hz.
21. Force at Negative Displacement
Question: k = 80 N/m and x = −0.05 m. Find F.
Show Solution
Given: k = 80 N/m, x = −0.05 m.
Formula: F = −kx.
Solution: F = −80(−0.05) = +4 N.
Final Answer: 4 N in positive direction, toward mean position.
22. Maximum Force
Question: k = 120 N/m and A = 0.05 m. Find maximum restoring force.
Show Solution
Given: k = 120 N/m, A = 0.05 m.
Formula: Fmax = kA.
Solution: Fmax = 120 × 0.05 = 6 N.
Final Answer: 6 N.
23. Speed at Half Amplitude
Question: Find speed at x = A/2 if vmax = 8 m/s.
Show Solution
Given: x = A/2, vmax = Aω = 8 m/s.
Formula: v = ω√(A² − x²).
Solution: v = Aω√(1 − 1/4) = vmax√3/2 = 4√3 m/s.
Final Answer: 4√3 m/s.
24. Acceleration at Half Amplitude
Question: Maximum acceleration is 16 m/s². Find acceleration magnitude at x = A/2.
Show Solution
Given: amax = ω²A = 16 m/s².
Formula: |a| = ω²x.
Solution: at x = A/2, |a| = ω²A/2 = 8 m/s².
Final Answer: 8 m/s².
25. Energy Ratio
Question: Find K:U at x = A/2.
Show Solution
Given: x = A/2.
Formula: U/E = x²/A², K/E = 1 − x²/A².
Solution: U/E = 1/4 and K/E = 3/4.
Final Answer: K:U = 3:1.
26. Amplitude From Energy
Question: E = 2 J and k = 100 N/m. Find amplitude.
Show Solution
Given: E = 2 J, k = 100 N/m.
Formula: E = 1/2 kA².
Solution: 2 = 50A², A² = 0.04, A = 0.20 m.
Final Answer: 0.20 m.
27. Number of Oscillations
Question: A particle has f = 2 Hz. How many oscillations in 30 s?
Show Solution
Given: f = 2 Hz, t = 30 s.
Formula: N = ft.
Solution: N = 2 × 30 = 60.
Final Answer: 60 oscillations.
28. Time For Oscillations
Question: Time period is 0.4 s. Find time for 25 oscillations.
Show Solution
Given: T = 0.4 s, N = 25.
Formula: total time = NT.
Solution: t = 25 × 0.4 = 10 s.
Final Answer: 10 s.
29. Mass Doubled
Question: In a spring-mass system, mass is doubled. What happens to time period?
Show Solution
Given: m' = 2m.
Formula: T = 2π√(m/k).
Solution: T' / T = √(2m/m) = √2.
Final Answer: Time period becomes √2 times.
30. Spring Constant Quadrupled
Question: If k becomes 4k, what happens to frequency?
Show Solution
Given: k' = 4k.
Formula: f = (1/2π)√(k/m).
Solution: f'/f = √4 = 2.
Final Answer: Frequency doubles.
31. Two Springs in Parallel
Question: Two springs k1 = 100 N/m and k2 = 300 N/m are in parallel with mass 1 kg. Find T.
Show Solution
Given: ktotal = k1 + k2 = 400 N/m, m = 1 kg.
Formula: T = 2π√(m/ktotal).
Solution: T = 2π√(1/400) = π/10 s.
Final Answer: π/10 s.
32. Two Springs in Series
Question: Two springs 200 N/m each are in series with 1 kg mass. Find T.
Show Solution
Given: k1 = k2 = 200 N/m.
Formula: 1/keq = 1/k1 + 1/k2.
Solution: keq = 100 N/m. T = 2π√(1/100) = π/5 s.
Final Answer: π/5 s.
33. Pendulum Length Quadrupled
Question: If pendulum length becomes 4L, what happens to T?
Show Solution
Given: L' = 4L.
Formula: T ∝ √L.
Solution: T'/T = √4 = 2.
Final Answer: Time period doubles.
34. Gravity Changes
Question: If g becomes g/4, what happens to pendulum time period?
Show Solution
Given: g' = g/4.
Formula: T ∝ 1/√g.
Solution: T'/T = √(g/g') = √4 = 2.
Final Answer: Time period doubles.
35. Phase Difference
Question: Two SHMs differ by half a cycle. Find phase difference.
Show Solution
Given: half cycle.
Formula: one full cycle = 2π rad.
Solution: phase difference = π rad.
Final Answer: π rad or 180°.
36. Projection of Circular Motion
Question: A particle moves in a circle of radius 0.2 m with angular speed 5 rad/s. Find amplitude and period of projection.
Show Solution
Given: radius = 0.2 m, ω = 5 rad/s.
Formula: A = radius, T = 2π/ω.
Solution: A = 0.2 m, T = 2π/5 s.
Final Answer: A = 0.2 m, T = 2π/5 s.
37. Find x When K = U
Question: At what displacement from mean position is kinetic energy equal to potential energy?
Show Solution
Given: K = U, total E = K + U.
Formula: U/E = x²/A².
Solution: If K = U, then U = E/2. So x²/A² = 1/2, hence x = A/√2.
Final Answer: x = A/√2 from mean position.
38. Displacement for Half Speed
Question: At what x is speed half of maximum speed?
Show Solution
Given: v = vmax/2.
Formula: v/vmax = √(1 − x²/A²).
Solution: 1/2 = √(1 − x²/A²). Squaring gives 1/4 = 1 − x²/A², so x²/A² = 3/4.
Final Answer: x = (√3/2)A.
39. Acceleration Ratio
Question: Find acceleration ratio at x = A/3 and x = 2A/3.
Show Solution
Given: x1 = A/3, x2 = 2A/3.
Formula: |a| = ω²x.
Solution: a1:a2 = x1:x2 = 1:2.
Final Answer: 1:2.
40. Combined Concept
Question: A 0.25 kg mass on a spring has A = 0.08 m and maximum speed 1.6 m/s. Find k.
Show Solution
Given: m = 0.25 kg, A = 0.08 m, vmax = 1.6 m/s.
Formula: vmax = Aω and ω² = k/m.
Solution: ω = 1.6/0.08 = 20 rad/s. k = mω² = 0.25 × 400 = 100 N/m.
Final Answer: 100 N/m.
Large PYQ and Practice Question Bank
This set mixes CBSE, NEET, JEE Main, JEE Advanced, IB, ICSE, IGCSE, A-Level, assertion-reason, true/false, conceptual, difficult numerical and case-study style questions. Answers are hidden for active recall.
1. Define periodic motion.
Write the definition and give two examples.
Show Answer
Motion that repeats itself after equal intervals of time is periodic motion. Examples: motion of Earth around Sun and hands of a clock.
2. Define oscillatory motion.
How is it different from general periodic motion?
Show Answer
Oscillatory motion is to-and-fro motion about a mean position. Uniform circular motion is periodic but not to-and-fro oscillatory motion.
3. What is amplitude?
Explain using a swing.
Show Answer
Amplitude is maximum displacement from mean position. For a swing, it is the maximum distance from its central hanging position.
4. State the condition for SHM.
Give the acceleration form.
Show Answer
Acceleration must be directly proportional to displacement and opposite in direction: a = −ω²x.
5. Explain the negative sign in F = −kx.
Why is it important?
Show Answer
The negative sign shows force is opposite to displacement and directed toward mean position.
6. If T = 0.5 s, find f.
Choose the correct value.
Show Answer
f = 1/T = 1/0.5 = 2 Hz.
7. Maximum velocity in SHM is found at which position?
Mean position or extreme position?
Show Answer
Maximum velocity is at mean position because kinetic energy is maximum there.
8. Maximum acceleration in SHM occurs where?
Explain briefly.
Show Answer
At extreme positions because |a| = ω²|x| and displacement magnitude is maximum there.
9. If amplitude doubles, what happens to total energy of spring SHM?
Assume k is constant.
Show Answer
E = 1/2 kA², so energy becomes four times.
10. Which graph is straight line: F-x or U-x?
For spring SHM.
Show Answer
F-x is a straight line with negative slope. U-x is a parabola.
11. Identify A and ω from x = 3 cos(5t).
x is in cm.
Show Answer
A = 3 cm and ω = 5 rad/s.
12. Find phase difference between x = A sin(ωt) and v = Aω cos(ωt).
State leading quantity.
Show Answer
Velocity leads displacement by π/2.
13. Projection of uniform circular motion is what?
Answer with condition.
Show Answer
Projection of uniform circular motion on a diameter is SHM.
14. If k/m = 25 s⁻², find ω.
Spring mass system.
Show Answer
ω = √(k/m) = 5 rad/s.
15. At x = A, what are v and acceleration magnitude?
Use SHM relations.
Show Answer
v = 0 and acceleration magnitude = ω²A.
16. When K = 3U, find x/A.
Energy based question.
Show Answer
U/E = 1/4, so x²/A² = 1/4. Hence x/A = 1/2 in magnitude.
17. If two identical springs are parallel, how does T change?
Compared with one spring.
Show Answer
k becomes 2k, so T becomes T/√2.
18. If two identical springs are series, how does T change?
Compared with one spring.
Show Answer
k becomes k/2, so T becomes √2 T.
19. Why is large-angle pendulum not exact SHM?
Conceptual trap.
Show Answer
Restoring torque contains sin θ. For large θ, sin θ is not approximately θ, so proportionality fails.
20. At what x is speed 1/√2 of maximum?
Find displacement magnitude.
Show Answer
v/vmax = √(1 − x²/A²) = 1/√2, so x = A/√2.
21. Describe energy transfer in SHM.
Use mean and extreme positions.
Show Answer
At mean position kinetic energy is maximum and potential energy minimum. At extremes potential energy is maximum and kinetic energy zero. Total energy remains constant in ideal SHM.
22. Explain phase in one sentence.
Use SHM language.
Show Answer
Phase specifies the stage of oscillation at a given time and decides position and direction of motion.
23. Give one example of periodic but non-oscillatory motion.
State reason.
Show Answer
Uniform circular motion of a fan blade is periodic but not to-and-fro about a mean position.
24. What is one complete oscillation?
Explain using a pendulum.
Show Answer
Motion from one extreme to the other and back to the original extreme with same direction of motion is one complete oscillation.
25. A pendulum completes 20 oscillations in 40 s. Find T.
Basic measurement.
Show Answer
T = total time/number = 40/20 = 2 s.
26. Why measure many oscillations in lab?
Practical skill.
Show Answer
Measuring many oscillations reduces percentage error in time measurement.
27. State the defining equation of SHM.
Use acceleration form.
Show Answer
a = −ω²x, where acceleration is proportional to displacement and directed toward equilibrium.
28. What does ω represent physically?
Not just unit.
Show Answer
ω is angular frequency, the rate at which phase changes with time, measured in rad/s.
29. Assertion: SHM is periodic. Reason: Acceleration is directed toward mean position and proportional to displacement.
Choose correct relation.
Show Answer
Both are true, and the reason correctly explains why ideal SHM repeats periodically.
30. Assertion: At mean position acceleration is maximum. Reason: Speed is maximum at mean position.
Evaluate.
Show Answer
Assertion is false. Reason is true. At mean position acceleration is zero, speed is maximum.
31. Every periodic motion is SHM.
True or false?
Show Answer
False. Example: uniform circular motion is periodic, but the circular motion itself is not SHM.
32. In SHM, force is maximum at extremes.
True or false?
Show Answer
True, because |F| = k|x| and |x| is maximum at extremes.
33. Why does a body cross the mean position instead of stopping there?
Explain in one paragraph.
Show Answer
At mean position the restoring force is zero, but the body has maximum speed due to conversion of potential energy into kinetic energy. Inertia carries it past the mean position.
34. Can acceleration and velocity be opposite in SHM?
Give an example location.
Show Answer
Yes. When the particle moves away from mean position, velocity is away from mean, while acceleration is toward mean.
35. Where is potential energy minimum?
For spring SHM.
Show Answer
At mean position, x = 0, so U = 1/2 kx² = 0 if zero is chosen at equilibrium.
36. A body has vmax = 3 m/s and amax = 12 m/s². Find A.
Use maximum relations.
Show Answer
vmax = Aω and amax = Aω². Therefore A = vmax²/amax = 9/12 = 0.75 m.
37. If E = 8 J and A = 0.2 m, find k.
Spring SHM.
Show Answer
E = 1/2 kA². So 8 = 1/2 k × 0.04 = 0.02k. Hence k = 400 N/m.
38. At x = A/3, find fraction of total energy that is kinetic.
Energy fraction.
Show Answer
U/E = x²/A² = 1/9. Therefore K/E = 8/9.
39. A pendulum clock is taken where g decreases. Will it gain or lose time?
Reason using period.
Show Answer
T ∝ 1/√g. If g decreases, T increases, so the clock runs slow and loses time.
40. If frequency is tripled, what happens to maximum acceleration for same amplitude?
Use amax = ω²A.
Show Answer
ω is proportional to f. If f triples, ω triples, so amax becomes 9 times for same A.
41. A student records 50 oscillations of a pendulum in 100 s.
Find T and state one source of error.
Show Answer
T = 100/50 = 2 s. A common error is reaction time while starting and stopping the stopwatch.
42. A spring block is pulled right and released from rest.
What are initial velocity and acceleration direction?
Show Answer
Initial velocity is zero. Acceleration is toward the mean position, opposite to displacement.
43. A tuning fork prong vibrates about its normal position.
Which concepts are involved?
Show Answer
Mean position, oscillatory motion, frequency, amplitude, restoring elastic force and energy exchange.
44. A fan rotates uniformly.
Is a point on a blade in SHM?
Show Answer
The point's circular motion is periodic, not SHM. Its projection on a diameter is SHM.
45. A pendulum is displaced by a very large angle.
Can small-angle SHM formula be used exactly?
Show Answer
No. The approximation sin θ ≈ θ fails for large angles, so the motion is not exact SHM.
46. Which quantity is zero at extreme position: velocity or acceleration?
Answer clearly.
Show Answer
Velocity is zero at extreme position. Acceleration is maximum in magnitude.
47. Which quantity is zero at mean position in SHM?
For ideal spring SHM.
Show Answer
Displacement, restoring force, acceleration and spring potential energy are zero at mean position.
48. If x is positive, what is the sign of acceleration?
Use a = −ω²x.
Show Answer
Acceleration is negative because it is opposite to positive displacement.
49. If displacement is negative, what is the direction of restoring force?
Use F = −kx.
Show Answer
F is positive, so the restoring force acts toward the positive direction, back toward mean position.
50. Name the formula connecting time period and angular frequency.
Give both forms.
Show Answer
ω = 2π/T and T = 2π/ω.
Revision Notes and Formula Sheet
Quick Revision Notes
- Periodic motion repeats after equal intervals.
- Oscillatory motion is to-and-fro about mean position.
- SHM requires a = −ω²x.
- Mean position has maximum speed.
- Extreme positions have maximum acceleration.
Formula Sheet
- T = 1/f
- f = 1/T
- ω = 2πf
- ω = 2π/T
- F = −kx
- a = −ω²x
- v = ω√(A² − x²)
- E = 1/2 kA²
Most Important Concepts
- Negative sign means restoring direction.
- Projection of uniform circular motion is SHM.
- Spring SHM: ω = √(k/m).
- Pendulum SHM is valid for small angles.
- Energy continuously changes form.
Exam Tips
- Check units before substitution.
- Convert cm to m in numerical problems.
- Use angular frequency in rad/s, not Hz.
- Draw a mean position line for direction questions.
- For energy ratios, use x²/A².
Common Mistakes
- Writing F = kx without direction.
- Calling every periodic motion SHM.
- Using pendulum formula for large angles.
- Assuming acceleration is maximum at mean.
- Confusing amplitude with total path length.
Last Day Revision Notes
- Memorize a = −ω²x as the definition.
- Mean: x = 0, F = 0, a = 0, v maximum.
- Extreme: x = ±A, v = 0, acceleration maximum.
- Spring: T = 2π√(m/k).
- Pendulum: T = 2π√(L/g).
If Periodic Motion or SHM is not clear and you are looking for a Physics Tutor, contact Kumar Sir.
Phone: +91-9958461445 Email: kumarsirphysics@gmail.com Website: https://kumarphysicsclasses.com