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The period of a simple pendulum inside a stationary lift is \(T\). The lift accelerates upwards with an acceleration of \(g / 3\). The time period of pendulum will be \(\quad\) (a) \(\sqrt{2} T\) (b) \(\frac{T}{\sqrt{2}}\) (c) \(\frac{\sqrt{3}}{2} T\) (d) \(\frac{T}{3}\)

Short Answer

Expert verified
The period of the pendulum is \( \frac{\sqrt{3}}{2} T \), which is option (c).

Step by step solution

01

Understanding the Pendulum Period Formula

The period of a simple pendulum is given by the formula \( T = 2 \pi \sqrt{\frac{L}{g}} \), where \( T \) is the period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.
02

Calculate Effective Gravity in the Lift

When the lift accelerates upwards with acceleration \( \frac{g}{3} \), the effective gravity becomes \( g_{\text{eff}} = g + \frac{g}{3} = \frac{4g}{3} \).
03

Determine New Period Formula in the Lift

Using the effective gravity, the new period \( T' \) of the pendulum is \( T' = 2 \pi \sqrt{\frac{L}{g_{\text{eff}}}} = 2 \pi \sqrt{\frac{L}{\frac{4g}{3}}} \).
04

Simplify the New Period Expression

The new period expression simplifies to \( T' = 2 \pi \sqrt{\frac{3L}{4g}} = \sqrt{\frac{3}{4}} \cdot 2 \pi \sqrt{\frac{L}{g}} \).
05

Relate New Period to Original Period

Since the original period is \( T = 2 \pi \sqrt{\frac{L}{g}} \), substituting this into the simplified expression for \( T' \) gives \( T' = \sqrt{\frac{3}{4}} T = \frac{\sqrt{3}}{2} T \).
06

Select the Correct Option

The correct option from the given choices is (c) \( \frac{\sqrt{3}}{2} T \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pendulum Period Formula
The period of a simple pendulum is fundamental to understanding how pendulums operate. This period, denoted as \( T \), represents the time taken for one complete cycle of the pendulum's swing from its starting point and back again. The formula to calculate this period is given by:\[T = 2 \pi \sqrt{\frac{L}{g}}\]Where:- \( L \) is the length of the pendulum.- \( g \) is the acceleration due to gravity (typically around \( 9.81 \, \text{m/s}^2 \) on Earth).This formula reveals how the period of a pendulum depends directly on the square root of its length and is inversely proportional to the square root of the gravitational acceleration. In essence, longer pendulums have longer periods, while stronger gravity results in shorter periods. Understanding this formula is crucial because it sets the stage for how the period changes when external factors, like acceleration, influence gravity.
Effective Gravity
When an object, like a pendulum, is within an accelerating frame (such as a lift), the apparent or "effective" gravity it experiences alters. In a lift accelerating upward, effective gravity increases because the pendulum feels both normal gravitational force and the upward acceleration of the lift. To calculate this effective gravity, you add the lift's acceleration to the typical gravitational acceleration:\[g_{\text{eff}} = g + a\]In this scenario, the lift ascends with an acceleration of \( \frac{g}{3} \), making the effective gravity:\[g_{\text{eff}} = g + \frac{g}{3} = \frac{4g}{3}\]This means the pendulum behaves as if it were in a location with one-third more gravity. Hence, effective gravity is vital to modify the original period formula correctly. The enhancement in gravity leads the pendulum to swing faster, affecting its period based on these altered dynamics.
Acceleration in a Lift
Acceleration impacts objects within a lift differently based on its direction. For a pendulum in such a scenario, understanding directional acceleration is key. When a lift accelerates upward, everything inside experiences increased effective gravity as if the gravitational pull has augmented. Consider the modified formula for the pendulum as a byproduct of this change:- The initial gravitational pull \( g \) is boosted by the upward acceleration \( \frac{g}{3} \).This modified environment now contributes to a change in the pendulum's period inside the accelerating lift:\[T' = 2 \pi \sqrt{\frac{L}{g_{\text{eff}}}} = 2 \pi \sqrt{\frac{L}{\frac{4g}{3}}}\]Upon simplification, we arrive at the new period:\[T' = \sqrt{\frac{3}{4}} T = \frac{\sqrt{3}}{2} T\]Thus, the time period decreases in an accelerated lift, affecting how often the pendulum swings back and forth. Recognizing the role of acceleration helps to intuitively understand these changes in the pendulum’s behavior, which might otherwise seem perplexing.

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Most popular questions from this chapter

A particle in SHM is described by the displacement function \(x(t)=A \cos (\omega t+\phi), \omega=2 \pi / T\). If the initial ( \(t=0\) ) position of the particle is \(1 \mathrm{~cm}\), its initial velocity is \(\pi \mathrm{cm} \mathrm{s}^{-1}\) and its angular frequency is \(\pi \mathrm{s}^{-1}\), then the amplitude of its motion is (a) \(\pi \mathrm{cm}\) (b) \(2 \mathrm{~cm}\) \(\begin{array}{ll}\text { (c) } \sqrt{2} \mathrm{~cm} & \text { (d) } 1 \mathrm{~cm}\end{array}\)

The acceleration due to gravity on the surface of moon is \(1.7 \mathrm{~m} / \mathrm{s}^{2}\). What is the time period of a simple pendulum on the surface of moon, if its time period on the surface of earth is \(3.5 \mathrm{~s}\) ? \(\left(g\right.\) on the surface of earth is \(9.8 \mathrm{~m} / \mathrm{s}^{2} .\) ) \(\quad\) [NCERT] (a) \(6.4 \mathrm{~s}\) (b) \(7.4 \mathrm{~s}\) (c) \(9.4 \mathrm{~s}\) (d) \(8.4 \mathrm{~s}\)

A particle moves in \(x y\)-plane according to the rule \(x=a \sin \omega t\) and \(y=a \cos \omega t\). The particle follows (a) an elliptical path (b) a circular path (c) a parabolic path (d) a straight line path inclined equally to \(x\) and \(y\)-axis

On a smooth inclined plane, a body of mass \(M\) is attached between two springs. The other ends of the springs are fixed to firm support. If each spring has force constant \(k\), the period of oscillation of the body (assuming the springs as massless) is (a) \(2 \pi[M / 2 k]^{1 / 2}\) (b) \(2 \pi[2 M / k]^{1 / 2}\) (c) \(2 \pi[M g \sin \theta / 2 k]^{1 / 2}\) (d) \(2 \pi[2 M g / k]^{1 / 2}\)

A particle is executing simple harmonic motion with an amplitude \(A\) and time period \(T\). The displacement of the particle after \(2 T\) period from its initial position is (a) \(\underline{A}\) (b) \(4 \mathrm{~A}\) (c) \(8 \mathrm{~A}\) (d) zero

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