If springs, pendulums or oscillations are not clear and you are looking for a Physics Tutor, contact Kumar Sir.Phone: +91-9958461445Email: kumarsirphysics@gmail.com

Call Kumar Sir WhatsApp Kumar Sir Visit Website Email Kumar Sir

CLASS 11 PHYSICS

Springs and Pendulums

Spring block systems, spring stiffness, time period, cut springs, series-parallel combinations, simple pendulum, seconds pendulum, effective length, numericals and PYQs.

CBSENEETJEE MainJEE AdvancedIBICSEIGCSEA-Level

1. Spring Block System

A spring block system is the most direct model of SHM. When a block attached to an ideal spring is displaced and released on a smooth horizontal surface, the spring provides restoring force.

Definition and Physical Meaning

If the spring is stretched or compressed by x, the spring force is F = −kx. The negative sign means the force is toward the mean position. The block accelerates back, crosses equilibrium due to inertia, and repeats the motion.

Example: a trolley attached to a spring on a frictionless track oscillates about the natural length position. At mean position speed is maximum; at extremes speed is zero.

Exam trap: the formula assumes ideal spring, no friction and small deformation within elastic limit.

Spring Block System

2. Spring Constant and Stiffness

Spring constant k measures stiffness. Higher spring constant means higher stiffness and a larger restoring force for the same displacement.

F = −kx

If k is large, even a small displacement produces a large restoring force. If k is small, the spring is soft and stretches more easily.

Exam Perspective

For the same mass, a stiffer spring oscillates faster because T = 2π√(m/k). Increasing k decreases T and increases frequency. Students often confuse stiffness with length; after cutting a spring, the smaller piece becomes stiffer.

Memory trick: tight spring, quick swing; soft spring, slow swing.

3. Time Period of Spring

Derivation

For a spring block, F = −kx and Newton's law gives ma = −kx, so a = −(k/m)x. Comparing with SHM form a = −ω²x gives ω = √(k/m).

T = 2π√(m/k)

m is mass attached to the spring and k is spring constant.

Meaning and Mistakes

The time period is independent of amplitude for ideal SHM. It increases with mass and decreases with stiffness.

Common mistake: using T = 2π√(k/m), which is inverted. Always remember heavy mass means longer time, so m must be in numerator.

4. Spring Cut into n Equal Parts

This is a very important NEET/JEE concept. If a spring of spring constant k is cut into n equal parts, each part becomes stiffer.

k' = nk

For one cut part, spring constant becomes n times the original spring constant.

T = 2π√(m/k)

Original spring time period.

T' = 2π√(m/(nk)) = T/√n

New time period using one cut part.

Reason: spring constant is inversely proportional to length for the same material and wire. Shorter spring means greater stiffness.

Spring Cut into n Equal Parts

Example 1

If a spring is cut into 2 equal parts, each part has k' = 2k. New period is T' = T/√2.

Example 2

If a spring is cut into 4 equal parts, each part has k' = 4k. New period is T' = T/2.

Example 3

If a spring is cut into 9 equal parts, each part has k' = 9k. New period is T' = T/3.

5. Springs in Series and Parallel

Series Combination

1/k_eq = 1/k₁ + 1/k₂

Series springs become softer. For two identical springs k and k, equivalent spring constant is k/2, so period increases by √2.

Parallel Combination

k_eq = k₁ + k₂

Parallel springs become stiffer. For two identical springs, equivalent spring constant is 2k, so period becomes T/√2.

Springs in Series

Springs in Parallel

6. Block Attached to Springs on Both Sides

When a block is connected between two springs and displaced, both springs provide restoring effect in the same direction toward equilibrium.

k_eff = k₁ + k₂

The two spring constants add because both springs oppose displacement.

T = 2π√(m/(k₁ + k₂))

Use this formula for a block attached to springs on both sides.

Exam trap: this is not series. The block feels restoring force from both sides, so effective stiffness is sum.

Block Between Two Springs

7. Simple Pendulum

Definition and Physical Meaning

A simple pendulum consists of a small heavy bob suspended by a light, inextensible string from a fixed support. For small angular displacement, its motion is approximately SHM.

The restoring component is due to gravity. When the bob is displaced, a component of weight acts tangentially toward the mean position.

Trap: large-angle pendulum is not exactly SHM because sin θ is not equal to θ.

Simple Pendulum

8. Time Period of Pendulum

T = 2π√(L/g)

L is effective length of pendulum and g is acceleration due to gravity.

For small angles, the restoring torque is proportional to angular displacement. This makes the pendulum approximately SHM.

Important Points

Time period does not depend on mass of bob.
Time period does not depend on small amplitude.
Increasing length increases time period.
Increasing g decreases time period.

Common mistake: using mass in pendulum period formula. Mass is absent.

9. Seconds Pendulum

Definition

A seconds pendulum has time period 2 seconds. It takes 1 second to go from one extreme to the other and 2 seconds for a complete oscillation.

T = 2 s

Near Earth's surface, its length is approximately 1 m because T = 2π√(L/g).

Seconds Pendulum

10. Effective Length

Effective length is the distance from the point of suspension to the centre of gravity of the bob.

L = distance from support to centre of bob

For a spherical bob, effective length is string length plus radius of bob if string length is measured up to the top of the bob.

Exam trap: do not use only string length unless it is clearly given as effective length.

Effective Length

11. Applications

Clocks

Pendulum clocks use nearly constant period for small amplitude oscillations. Temperature changes can alter length and affect timekeeping.

Vehicle Suspension

Springs in vehicles store and release energy, reducing shocks. Stiffer springs give higher natural frequency.

Seismometers

Spring-mass systems help detect ground vibrations. Understanding natural frequency prevents wrong readings.

Measuring g

Using T = 2π√(L/g), a pendulum experiment can estimate local gravitational acceleration.

Engineering Design

Machines use springs to control vibration and avoid resonance at operating frequencies.

Exam Use

Spring and pendulum formulas appear in CBSE derivations, NEET direct questions and JEE combination problems.

Need Help With Springs or Pendulums?

If springs, pendulums or oscillations are not clear and you are looking for a Physics Tutor, contact Kumar Sir.

Call Kumar Sir WhatsApp Kumar Sir Visit Website Email Kumar Sir

Formula Sheet

T = 2π√(m/k)

Spring block period.

ω = √(k/m)

Spring angular frequency.

k' = nk

Spring cut into n equal parts.

T' = T/√n

Using one cut part.

1/k_eq = 1/k₁ + 1/k₂

Series springs.

k_eq = k₁ + k₂

Parallel or both-side springs.

T = 2π√(m/(k₁ + k₂))

Block attached between two springs.

T = 2π√(L/g)

Simple pendulum period.

T = 2 s

Seconds pendulum.

12. 40 Solved Numericals

CBSE Easy

1. Spring period

Question: m = 1 kg and k = 100 N/m. Find T.

Show Solution

Given: m = 1, k = 100.

Formula: T = 2π√(m/k).

Solution: T = 2π√(1/100) = π/5 s.

Final Answer: π/5 s.

CBSE Easy

2. Angular frequency

Question: k = 200 N/m, m = 2 kg. Find ω.

Show Solution

Given: k = 200, m = 2.

Formula: ω = √(k/m).

Solution: ω = √100 = 10 rad/s.

Final Answer: 10 rad/s.

NEET Easy

3. Effect of mass

Question: Mass is made 4 times. What happens to spring period?

Show Solution

Given: m' = 4m.

Formula: T ∝ √m.

Solution: T' = 2T.

Final Answer: Period doubles.

NEET Easy

4. Effect of stiffness

Question: k becomes 4k. What happens to T?

Show Solution

Given: k' = 4k.

Formula: T ∝ 1/√k.

Solution: T' = T/2.

Final Answer: Period halves.

JEE Main Medium

5. Cut into 4 parts

Question: A spring is cut into 4 equal parts. Find k' for each part.

Show Solution

Given: n = 4.

Formula: k' = nk.

Solution: k' = 4k.

Final Answer: 4k.

JEE Main Medium

6. Period of cut spring

Question: Original period is 6 s. Spring is cut into 9 parts and one part is used. Find new T.

Show Solution

Given: T = 6 s, n = 9.

Formula: T' = T/√n.

Solution: T' = 6/3 = 2 s.

Final Answer: 2 s.

JEE Advanced Medium

7. Cut into 2 parts

Question: Original period is T. One half spring is used. Find new period.

Show Solution

Given: n = 2.

Formula: T' = T/√n.

Solution: T' = T/√2.

Final Answer: T/√2.

JEE Advanced Medium

8. Cut into 3 parts

Question: k = 60 N/m. Spring is cut into 3 equal parts. Find each k'.

Show Solution

Given: k = 60, n = 3.

Formula: k' = nk.

Solution: k' = 180 N/m.

Final Answer: 180 N/m.

CBSE Medium

9. Series identical springs

Question: Two springs each k are in series. Find k_eq.

Show Solution

Given: k and k in series.

Formula: 1/k_eq = 1/k + 1/k.

Solution: k_eq = k/2.

Final Answer: k/2.

NEET Medium

10. Parallel identical springs

Question: Two springs each k are in parallel. Find k_eq.

Show Solution

Given: k and k in parallel.

Formula: k_eq = k + k.

Solution: k_eq = 2k.

Final Answer: 2k.

JEE Main Medium

11. Series numerical

Question: k₁ = 100 N/m, k₂ = 100 N/m in series, m = 2 kg. Find T.

Show Solution

Given: k_eq = 50 N/m, m = 2.

Formula: T = 2π√(m/k_eq).

Solution: T = 2π√(2/50) = 2π/5 s.

Final Answer: 2π/5 s.

JEE Main Medium

12. Parallel numerical

Question: k₁ = 100, k₂ = 300 N/m, m = 1 kg. Find T.

Show Solution

Given: k_eq = 400 N/m.

Formula: T = 2π√(m/k_eq).

Solution: T = 2π√(1/400) = π/10 s.

Final Answer: π/10 s.

NEET Medium

13. Both side springs

Question: k₁ = 50, k₂ = 150 N/m, m = 2 kg. Find T.

Show Solution

Given: k_eff = 200, m = 2.

Formula: T = 2π√(m/(k₁ + k₂)).

Solution: T = 2π√(2/200) = π/5 s.

Final Answer: π/5 s.

CBSE Easy

14. Pendulum period

Question: L = 1 m, g = π² m/s². Find T.

Show Solution

Given: L = 1, g = π².

Formula: T = 2π√(L/g).

Solution: T = 2π√(1/π²) = 2 s.

Final Answer: 2 s.

CBSE Easy

15. Seconds pendulum length

Question: T = 2 s and g = π². Find L.

Show Solution

Given: T = 2, g = π².

Formula: T = 2π√(L/g).

Solution: 2 = 2π√(L/π²), so L = 1 m.

Final Answer: 1 m.

NEET Medium

16. Length quadrupled

Question: Pendulum length becomes 4L. What happens to T?

Show Solution

Given: L' = 4L.

Formula: T ∝ √L.

Solution: T' = 2T.

Final Answer: Doubles.

NEET Medium

17. g becomes g/4

Question: If g becomes g/4, what happens to T?

Show Solution

Given: g' = g/4.

Formula: T ∝ 1/√g.

Solution: T' = 2T.

Final Answer: Doubles.

JEE Main Medium

18. Find g

Question: L = 1 m, T = 2 s. Find g.

Show Solution

Given: L = 1, T = 2.

Formula: g = 4π²L/T².

Solution: g = 4π²/4 = π² m/s².

Final Answer: π² m/s².

JEE Advanced Medium

19. Effective length

Question: String length to top of spherical bob is 0.9 m and bob radius is 0.1 m. Find effective length.

Show Solution

Given: string to top = 0.9 m, radius = 0.1 m.

Formula: L = string length + radius.

Solution: L = 1.0 m.

Final Answer: 1.0 m.

JEE Advanced Medium

20. Pendulum period from effective length

Question: L = 0.25 m, g = π². Find T.

Show Solution

Given: L = 0.25, g = π².

Formula: T = 2π√(L/g).

Solution: T = 2π√(0.25/π²) = 1 s.

Final Answer: 1 s.

CBSE Medium

21. Mass independence

Question: Pendulum bob mass is doubled. What happens to T?

Show Solution

Given: mass changes.

Formula: T = 2π√(L/g).

Solution: mass is absent.

Final Answer: No change.

NEET Medium

22. Frequency of spring

Question: k = 400 N/m, m = 1 kg. Find f.

Show Solution

Given: ω = √400 = 20 rad/s.

Formula: f = ω/(2π).

Solution: f = 10/π Hz.

Final Answer: 10/π Hz.

JEE Main Difficult

23. Three identical series springs

Question: Three identical springs k are in series. Find period compared to one spring.

Show Solution

Given: k_eq = k/3.

Formula: T ∝ 1/√k.

Solution: T' = √3 T.

Final Answer: √3 times.

JEE Main Difficult

24. Three identical parallel springs

Question: Three identical springs k are in parallel. Find period compared to one spring.

Show Solution

Given: k_eq = 3k.

Formula: T ∝ 1/√k.

Solution: T' = T/√3.

Final Answer: T/√3.

JEE Advanced Difficult

25. Cut and mass changed

Question: Spring cut into 4 parts and mass becomes 4m. One part is used. Find new period compared to original.

Show Solution

Given: k' = 4k, m' = 4m.

Formula: T' = 2π√(m'/k').

Solution: T' = 2π√(4m/4k) = T.

Final Answer: No change.

JEE Advanced Difficult

26. Both side identical springs

Question: A block between two identical springs k has period?

Show Solution

Given: k_eff = 2k.

Formula: T = 2π√(m/2k).

Solution: substitute effective stiffness.

Final Answer: 2π√(m/2k).

CBSE Medium

27. Extension force

Question: k = 50 N/m, x = 0.2 m. Find restoring force magnitude.

Show Solution

Given: k = 50, x = 0.2.

Formula: F = kx.

Solution: F = 10 N.

Final Answer: 10 N toward mean.

NEET Medium

28. Static extension method

Question: A mass stretches spring by y at equilibrium. Find period.

Show Solution

Given: mg = ky.

Formula: T = 2π√(m/k).

Solution: m/k = y/g.

Final Answer: T = 2π√(y/g).

JEE Main Difficult

29. Static extension numerical

Question: Static extension is 0.25 m. Take g = π². Find T.

Show Solution

Given: y = 0.25, g = π².

Formula: T = 2π√(y/g).

Solution: T = 2π√(0.25/π²) = 1 s.

Final Answer: 1 s.

JEE Advanced Difficult

30. Series unequal springs

Question: k₁ = 100, k₂ = 300 N/m. Find series k_eq.

Show Solution

Given: two springs in series.

Formula: k_eq = k₁k₂/(k₁ + k₂).

Solution: k_eq = 30000/400 = 75 N/m.

Final Answer: 75 N/m.

CBSE Easy

31. Seconds pendulum half swing

Question: Seconds pendulum takes how much time from one extreme to other?

Show Solution

Given: T = 2 s.

Formula: extreme to opposite extreme = T/2.

Solution: time = 1 s.

Final Answer: 1 s.

NEET Easy

32. Pendulum on Moon

Question: g decreases. What happens to pendulum period?

Show Solution

Given: lower g.

Formula: T ∝ 1/√g.

Solution: T increases.

Final Answer: Period increases.

JEE Main Medium

33. Clock slow or fast

Question: Pendulum length increases. Clock runs?

Show Solution

Given: L increases.

Formula: T ∝ √L.

Solution: T increases, fewer oscillations per time.

Final Answer: Clock runs slow.

JEE Advanced Medium

34. Pendulum length for T = 4 s

Question: If g = π², find L for T = 4 s.

Show Solution

Given: T = 4, g = π².

Formula: L = gT²/(4π²).

Solution: L = π² × 16/(4π²) = 4 m.

Final Answer: 4 m.

NEET Medium

35. Spring period from frequency

Question: Spring frequency is 5 Hz. Find period.

Show Solution

Given: f = 5 Hz.

Formula: T = 1/f.

Solution: T = 0.2 s.

Final Answer: 0.2 s.

CBSE Medium

36. Find mass

Question: k = 100 N/m, T = 2π/5 s. Find m.

Show Solution

Given: k = 100, T = 2π/5.

Formula: T = 2π√(m/k).

Solution: √(m/100) = 1/5, so m = 4 kg.

Final Answer: 4 kg.

JEE Main Difficult

37. Find k from period

Question: m = 2 kg, T = π s. Find k.

Show Solution

Given: m = 2, T = π.

Formula: T = 2π√(m/k).

Solution: 1/2 = √(2/k), so k = 8 N/m.

Final Answer: 8 N/m.

JEE Advanced Difficult

38. Cut part with k value

Question: Original k = 40 N/m. Cut into 5 parts. Mass 2 kg attached to one part. Find ω.

Show Solution

Given: k' = 5k = 200 N/m, m = 2.

Formula: ω = √(k'/m).

Solution: ω = √100 = 10 rad/s.

Final Answer: 10 rad/s.

JEE Advanced Difficult

39. Both side unequal

Question: k₁ = k, k₂ = 3k. Find period.

Show Solution

Given: k_eff = 4k.

Formula: T = 2π√(m/k_eff).

Solution: T = 2π√(m/4k) = π√(m/k).

Final Answer: π√(m/k).

JEE Advanced Difficult

40. Compare spring and pendulum

Question: Spring block has T = 2π√(m/k). Pendulum has same period. Find L.

Show Solution

Given: 2π√(L/g) = 2π√(m/k).

Formula: equate square roots.

Solution: L/g = m/k, so L = mg/k.

Final Answer: L = mg/k.

13. 50 PYQs and Exam Questions

CBSE

1. State time period of spring block.

Write the formula.

Show Answer

T = 2π√(m/k).

CBSE

2. What is spring constant?

Define stiffness.

Show Answer

Spring constant is restoring force per unit displacement. Higher k means greater stiffness.

CBSE

3. Does spring period depend on amplitude?

Ideal spring.

Show Answer

No, for ideal SHM.

CBSE

4. State pendulum time period.

Small oscillations.

Show Answer

T = 2π√(L/g).

CBSE

5. What is effective length?

For pendulum.

Show Answer

Distance from point of suspension to centre of gravity of bob.

NEET

6. Spring cut into n parts.

What is k'?

Show Answer

Each part has k' = nk.

NEET

7. Spring cut into n parts.

What is T' for one part?

Show Answer

T' = T/√n.

NEET

8. Two springs in parallel.

Equivalent k?

Show Answer

k_eq = k₁ + k₂.

NEET

9. Two springs in series.

Equivalent k?

Show Answer

1/k_eq = 1/k₁ + 1/k₂.

NEET

10. Both-side springs.

Effective k?

Show Answer

k_eff = k₁ + k₂.

JEE Main

11. If k is quadrupled, T becomes?

Same mass.

Show Answer

T becomes T/2.

JEE Main

12. If mass is quadrupled, T becomes?

Same spring.

Show Answer

T becomes 2T.

JEE Main

13. Seconds pendulum period?

Definition.

Show Answer

2 seconds.

JEE Main

14. Seconds pendulum length near Earth?

Approximate.

Show Answer

Approximately 1 m.

JEE Main

15. Does pendulum period depend on bob mass?

Small oscillations.

Show Answer

No.

JEE Advanced

16. Why small-angle condition?

Pendulum SHM.

Show Answer

Because sin θ ≈ θ only for small angles, giving restoring torque proportional to displacement.

JEE Advanced

17. Cut spring and stiffness.

Why does k increase?

Show Answer

For same spring material, k is inversely proportional to length; shorter part is stiffer.

JEE Advanced

18. Block between springs formula.

Write T.

Show Answer

T = 2π√(m/(k₁ + k₂)).

JEE Advanced

19. Static extension formula.

Vertical spring period.

Show Answer

T = 2π√(y/g), where y is static extension.

JEE Advanced

20. Parallel springs effect.

Soft or stiff?

Show Answer

Stiffer; equivalent k increases.

21. Explain restoring force in spring.

Conceptual.

Show Answer

It acts opposite to displacement and pulls/pushes block toward equilibrium.

22. Explain pendulum restoring force.

Conceptual.

Show Answer

The tangential component of weight acts toward the mean position.

ICSE

23. Give use of seconds pendulum.

Application.

Show Answer

It is used in timekeeping and pendulum clocks.

ICSE

24. What happens if pendulum length increases?

Period effect.

Show Answer

Time period increases.

IGCSE

25. How to reduce timing error in pendulum lab?

Practical.

Show Answer

Measure time for many oscillations and divide by number of oscillations.

IGCSE

26. What is one oscillation?

Pendulum.

Show Answer

Motion from one extreme to the other and back to original extreme.

A-Level

27. Derive spring angular frequency.

State result.

Show Answer

ma = −kx gives a = −(k/m)x, so ω = √(k/m).

A-Level

28. Derive pendulum period condition.

Small angle.

Show Answer

For small θ, restoring acceleration is proportional to displacement, giving T = 2π√(L/g).

Assertion-Reason

29. Assertion: A cut spring is stiffer. Reason: spring constant is inversely proportional to length.

Evaluate.

Show Answer

Both are true and reason explains assertion.

Assertion-Reason

30. Assertion: Pendulum period depends on mass. Reason: heavier bob has larger weight.

Evaluate.

Show Answer

Assertion is false. Reason is true but not relevant; mass cancels in pendulum motion.

True/False

31. Higher k means higher stiffness.

True or false?

Show Answer

True.

True/False

32. Series springs become stiffer.

True or false?

Show Answer

False; series springs become softer.

True/False

33. Parallel springs become stiffer.

True or false?

Show Answer

True.

True/False

34. Seconds pendulum has period 1 s.

True or false?

Show Answer

False; its period is 2 s.

Conceptual

35. Why does both-side spring k add?

Explain.

Show Answer

Both springs exert restoring forces toward equilibrium, so total restoring force is −(k₁ + k₂)x.

Conceptual

36. Why is pendulum not exact SHM at large angle?

Explain.

Show Answer

Because restoring term contains sin θ, not θ; proportionality fails for large θ.

Conceptual

37. Why does softer spring have larger period?

Explain.

Show Answer

Smaller k gives weaker restoring force and smaller acceleration, so oscillation is slower.

Conceptual

38. Why does longer pendulum have larger period?

Explain.

Show Answer

Longer pendulum has smaller angular acceleration for same angular displacement, so it oscillates slower.

Difficult

39. Spring cut into 16 parts.

New period with one part?

Show Answer

T' = T/4.

Difficult

40. Four identical springs parallel.

New period compared to one spring?

Show Answer

T' = T/2.

Difficult

41. Four identical springs series.

New period compared to one spring?

Show Answer

T' = 2T.

Difficult

42. Pendulum clock taken to lower g.

Fast or slow?

Show Answer

It runs slow because period increases.

Case Study

43. A vehicle uses stiff springs.

What happens to natural frequency?

Show Answer

Natural frequency increases because ω = √(k/m).

Case Study

44. A pendulum bob is replaced by heavier bob.

What happens to period?

Show Answer

No change for small oscillations.

Case Study

45. A spring is cut shorter.

What happens to stiffness?

Show Answer

Stiffness increases.

Case Study

46. Two side springs pull block back.

Equivalent spring constant?

Show Answer

k₁ + k₂.

Graph/Concept

47. Spring F-x graph slope?

Meaning.

Show Answer

Slope magnitude is spring constant k.

Graph/Concept

48. Pendulum period vs √L graph?

Shape.

Show Answer

Straight line because T ∝ √L.

Graph/Concept

49. Spring period vs √m graph?

Shape.

Show Answer

Straight line because T ∝ √m.

Graph/Concept

50. Spring period vs 1/√k graph?

Shape.

Show Answer

Straight line because T ∝ 1/√k.

14. Quick Revision Notes

Spring Block

F = −kx
ω = √(k/m)
T = 2π√(m/k)
Higher k means higher stiffness.

Cut Spring

Cut into n equal parts.
Each part: k' = nk.
Using one part: T' = T/√n.
Shorter spring is stiffer.

Combinations

Series: reciprocal formula.
Parallel: k values add.
Both-side springs: k_eff = k₁ + k₂.

Pendulum

T = 2π√(L/g)
Valid for small angles.
Mass independent.
Seconds pendulum: T = 2 s.

Exam Tips

Check whether springs are series, parallel or both-side.
Use effective length, not always string length.
For cut spring, never write k/n.

Last Day Notes

Heavy mass: larger spring period.
Stiffer spring: smaller period.
Longer pendulum: larger period.
Lower g: larger pendulum period.

If springs, pendulums or oscillations are not clear and you are looking for a Physics Tutor, contact Kumar Sir.

Phone: +91-9958461445 Email: kumarsirphysics@gmail.com Website: https://kumarphysicsclasses.com

Call Kumar Sir WhatsApp Kumar Sir Email Kumar Sir Visit Website