physics tutor in breach candy mumbai

Physics Tutor in Breach Candy Mumbai for NEET, IIT JEE, CBSE, IB, IGCSE, Edexcel, A-Level and AP Physics with pendulum in liquid derivation

Physics Tutor in Breach Candy Mumbai Call  WhatsApp: +91-9958461445

If you live in Breach Candy Mumbai and you are preparing for NEET, IIT JEE, CBSE, IB, IGCSE, Edexcel, A-Level or AP Physics, then Physics can sometimes become very confusing, especially in chapters like Simple Harmonic Motion.

Breach Candy is one of South Mumbai’s upscale residential localities near the Arabian Sea, also connected with areas like Cumballa Hill, Warden Road / Bhulabhai Desai Road, Peddar Road, Altamount Road and Malabar Hill. (Mygate)

Now take this SHM question:

A particle executing SHM of amplitude A has displacement A/2 at time T/4 and negative velocity. Find the epoch of the particle.

Here, epoch means initial phase angle, usually written as ϕ.

We start with:

y = A sin(ωt + ϕ)

Differentiate it:

v = Aω cos(ωt + ϕ)

Given:

y = A/2

So,

A/2 = A sin(ωt + ϕ)

Therefore,

sin(ωt + ϕ) = 1/2

At t = T/4,

ωt = (2π/T) × (T/4) = π/2

So,

sin(π/2 + ϕ) = 1/2

But velocity is negative, so:

v = Aω cos(π/2 + ϕ) < 0

That means the angle must be in the second quadrant where sine is positive and cosine is negative.

So,

π/2 + ϕ = 5π/6

Therefore,

ϕ = 5π/6 − π/2 = π/3

So the epoch of the particle is:

π/3 radian

This is exactly the type of question where many students get stuck. They know the formula, but they do not know how to use the sign of velocity. Kumar Sir explains these questions step by step: first displacement equation, then velocity equation, then quadrant selection, then final phase angle.

If you are stuck in SHM, waves, mechanics, electrostatics, current electricity, optics or modern physics, you can contact Kumar Sir for online Physics classes.

Call / WhatsApp: +91-9958461445
Email: kumarsirphysics@gmail.com
Website: Kumar Physics Classes

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Effect on Time Period of Simple Pendulum

In Simple Harmonic Motion, the time period of a simple pendulum depends mainly on the effective length of the pendulum and the effective acceleration due to gravity. For small oscillations, the standard formula is:

T = 2π√(l / g)

Here, l is the effective length of the pendulum and g is the effective acceleration due to gravity. Whenever length increases, time period increases. Whenever effective gravity decreases, time period increases.

Important Conditions Affecting Time Period

  • 1. Length increases: Time period increases because T ∝ √l.
  • 2. Length decreases: Time period decreases.
  • 3. Gravity increases: Time period decreases because T ∝ 1/√g.
  • 4. Gravity decreases: Time period increases.
  • 5. Mass of bob changes: No effect on time period.
  • 6. Amplitude changes slightly: For small oscillations, no effect.
  • 7. Large amplitude: Time period slightly increases.
  • 8. Temperature increases: Length of pendulum increases due to expansion, so time period increases.
  • 9. Temperature decreases: Length decreases, so time period decreases.

Second Pendulum

A second pendulum is a pendulum whose time period is exactly 2 seconds. It takes 1 second to go from one extreme to the other and 2 seconds to complete one full oscillation.

T = 2 s

For a second pendulum on Earth, its length is nearly 1 metre. More accurately, it is about 99.3 cm.

Effect of Temperature on Second Pendulum

If temperature increases, the metallic rod or string expands. Its length increases. Since T = 2π√(l/g), the time period increases. Therefore, the clock becomes slow because each oscillation takes more time.

T' = T(1 + αΔθ / 2)

Here, α is the coefficient of linear expansion and Δθ is the rise in temperature.

Pendulum Taken to Height Above Earth

When a pendulum is taken above the surface of Earth, the value of g decreases. Since time period is inversely proportional to √g, the time period increases.

g' = g(1 - 2h/R)
T' = T(1 + h/R)

So, at height, the pendulum clock becomes slow.

Pendulum Taken Deep Inside Earth

When a pendulum is taken inside the Earth, the value of g decreases with depth. Therefore, time period increases.

g' = g(1 - d/R)
T' = T / √(1 - d/R)

At the centre of Earth, g becomes zero. Therefore, the time period becomes infinite and the pendulum will not oscillate.

Pendulum Taken to Infinity

At infinity, gravitational field becomes almost zero. Therefore, restoring force becomes zero and the pendulum cannot perform normal SHM.

g → 0, therefore T → ∞

Pendulum in a Lift

  • Lift moving upward with acceleration a: effective gravity = g + a, so time period decreases.
  • Lift moving downward with acceleration a: effective gravity = g - a, so time period increases.
  • Lift falling freely: effective gravity = 0, so time period becomes infinite.
T = 2π√(l / g effective)

Pendulum in an Accelerating Vehicle

If the support of pendulum accelerates horizontally with acceleration a, effective gravity becomes the resultant of g and a.

g effective = √(g² + a²)

So the time period decreases because effective gravity increases.

Pendulum in Electric Field

If the bob is charged and placed in an electric field, electric force changes the effective acceleration. Depending on the direction of electric force, time period may increase or decrease.

g effective = g ± qE/m

Pendulum on Moon

On the Moon, gravity is nearly g/6. Therefore, time period increases.

T moon = √6 × T earth

Ball Dropped Through a Tunnel Inside Earth

If a straight tunnel is made through the Earth and a ball is dropped inside it, the ball performs SHM because the restoring force is proportional to displacement from the centre of Earth.

F ∝ -x
T = 2π√(R / g)

This time period is approximately 84.6 minutes. This is a very important result for NEET and IIT JEE.

Complete Summary

  • Time period increases if length increases.
  • Time period decreases if length decreases.
  • Time period increases if gravity decreases.
  • Time period decreases if gravity increases.
  • Temperature rise increases time period.
  • At height, time period increases.
  • At depth, time period increases.
  • At Earth’s centre, pendulum stops oscillating.
  • At infinity, time period becomes infinite.
  • In upward accelerating lift, time period decreases.
  • In downward accelerating lift, time period increases.
  • In freely falling lift, time period becomes infinite.
  • On Moon, time period increases.
  • For large amplitude, time period slightly increases.
  • Mass of bob has no effect.
  • For a ball oscillating through Earth tunnel, T = 2π√(R/g).

For NEET, IIT JEE, CBSE, IB, IGCSE, Edexcel, A-Level and AP Physics, these pendulum time period cases are very important. If a student understands effective gravity and effective length, almost every pendulum question becomes easy.

SHM of Floating Body and Pendulum in Liquid

In Physics, many oscillation problems become easy when we understand the idea of restoring force. If a body is floating in a liquid and it is slightly pushed downward or upward, then the upthrust changes. This change in upthrust provides the restoring force and the body performs Simple Harmonic Motion.

1. Floating Body Performing SHM in Liquid

Suppose a body of density σ is floating in a liquid of density ρ. Let the area of cross-section of the body be A. If the body is slightly pushed down by a distance x, then extra volume of liquid displaced will be:

Extra Volume = A x

Extra upthrust acting on the body will be:

Restoring Force = ρ A x g

This restoring force acts upward, opposite to displacement. Therefore:

F = -ρ A g x

For SHM:

F = -mω²x

Comparing both:

mω² = ρ A g

Now mass of the body:

m = σ V

If the body has height h and cross-sectional area A, then:

V = A h
m = σ A h

So:

σ A h ω² = ρ A g
ω² = ρ g / σ h

Therefore, the time period is:

T = 2π√(σh / ρg)

This is the time period of vertical oscillation of a floating body in a liquid. Here, σ is density of body, ρ is density of liquid, and h is the height of the body.

Important Special Result

If the body is floating, then at equilibrium:

Weight of body = Upthrust
σ A h g = ρ A y g

Here, y is the submerged depth of the body at equilibrium.

y = σh / ρ

So time period can also be written as:

T = 2π√(y / g)

This is a very beautiful result. The time period depends on the submerged depth at equilibrium.

2. Simple Pendulum in Air and Then in Liquid

For a simple pendulum in air, the time period is:

T = 2π√(l / g)

Now suppose the pendulum bob is placed inside a liquid of density ρ. Let the density of bob be σ. Due to buoyancy, the effective weight of the bob decreases.

Actual weight of bob:

W = Vσg

Buoyant force:

B = Vρg

Effective weight:

W' = Vσg - Vρg
W' = Vg(σ - ρ)

But effective weight can also be written as:

W' = mg'

Since mass of bob is:

m = Vσ

Therefore:

Vσg' = Vg(σ - ρ)
g' = g(σ - ρ) / σ
g' = g(1 - ρ/σ)

So, in liquid, the effective acceleration due to gravity becomes:

g effective = g(1 - ρ/σ)

Therefore, the time period of pendulum in liquid becomes:

T liquid = 2π√[ l / { g(1 - ρ/σ) } ]

If time period in air is:

T air = 2π√(l / g)

Then:

T liquid = T air / √(1 - ρ/σ)

Final Comparison

Since 1 - ρ/σ is less than 1, the time period in liquid becomes greater than the time period in air.

T liquid > T air

So, when a pendulum is taken from air into water or any liquid, its time period increases because buoyancy reduces effective gravity.

Important Notes for NEET and IIT JEE

If ρ = 0, then liquid effect is zero and the formula becomes the normal air formula.

T = 2π√(l / g)

If ρ = σ, then effective gravity becomes zero. In that case, the pendulum will not oscillate normally and time period becomes infinite.

ρ = σ ⇒ g effective = 0 ⇒ T = ∞

If ρ > σ, the bob cannot remain properly submerged as a normal pendulum bob because buoyant force becomes greater than weight.

These two derivations are very important for NEET Physics, IIT JEE Physics, CBSE Class 11 Physics, A-Level Physics and AP Physics. Once the student understands restoring force and effective gravity, these questions become very easy.

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