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Chapter 15: Problem 2

Two pendulums of length \(121 \mathrm{~cm}\) and \(100 \mathrm{~cm}\) start vibrating. At same instant the two are in the mean position in the same phase. After how many vibrations of the shorter pendulum, the two will be in phase at the mean position? (a) 10 (b) 11 (c) 20 (d) 21

Short Answer

Expert verified
The shorter pendulum makes 10 vibrations.

Step by step solution

01

Calculate Individual Time Periods

First, find the time periods of the two pendulums using the formula for the period of a simple pendulum: \( T = 2 \pi \sqrt{\frac{L}{g}} \), where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity (approximately \( 9.8 \ m/s^2 \)). We have: - For the first pendulum with length \( 121 \, \text{cm} \): \[ T_1 = 2 \pi \sqrt{\frac{121}{980}} \] - For the second pendulum with length \( 100 \, \text{cm} \): \[ T_2 = 2 \pi \sqrt{\frac{100}{980}} \].
02

Determine the Ratio of Time Periods

Once we have the expressions for the time periods, simplify the ratios: \[ \frac{T_1}{T_2} = \frac{\sqrt{\frac{121}{980}}}{\sqrt{\frac{100}{980}}} = \frac{\sqrt{121}}{\sqrt{100}} = \frac{11}{10} \].
03

Find the Ratio of Vibrations to Bring Them Into Phase

For the two pendulums to be in phase again at the mean position, the ratio of the number of vibrations should be the reciprocal of the period ratio. Hence, if \( n \) is the number of vibrations for the shorter pendulum, we have:\[ \frac{n}{n+1} = \frac{10}{11} \].
04

Solve the Equation

Set up the equation from Step 3: \[ 11n = 10(n + 1) \] Solving this, we simplify the equation to find \( n \): \[ 11n = 10n + 10 \] \[ n = 10 \].

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Key Concepts

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

Time Period Calculation
The time period calculation of a simple pendulum is crucial to understanding pendulum motion. It helps in determining how long it takes for the pendulum to complete one full cycle of swinging back and forth. The formula to calculate the time period is
  • \(T = 2\pi \sqrt{\frac{L}{g}}\)
In this formula:
  • \(T\) represents the time period of the pendulum,
  • \(L\) stands for the length of the pendulum, and
  • \(g\) denotes the acceleration due to gravity, which is approximately \(9.8 \mathrm{~m/s^2}\).
For example, in the given exercise, we have pendulums with lengths of \(121\) cm and \(100\) cm. By substituting these lengths into the formula, you can calculate their respective time periods. These calculations help to understand how longer pendulums take more time to complete a swing compared to shorter ones, assuming gravity remains constant. Such calculations are fundamental for predicting pendulum movements.
Pendulum Length and Vibration
A pendulum's length directly affects its time period and, consequently, its vibrations. Simply put:
  • Longer pendulums take more time to complete a vibration.
  • Shorter pendulums vibrate more quickly.
This relationship, described by the time period formula, is due to how the pendulum's length influences the arc through which it swings. Consider two pendulums, as in the exercise, where one has a length of \(121\) cm, and the other is \(100\) cm. Calculating their time periods:
  • The longer pendulum will have a larger time period \(T_1\).
  • The shorter pendulum, with a length of \(100\) cm, gives a smaller time period \(T_2\).
These different time periods mean they will not always swing in phase together, meaning they don't always reach their highest or mean positions simultaneously. Understanding this helps when analyzing synchronized motions like swings or building clocks.
Phase Relationship in Oscillations
Phase relationship in oscillations refers to the relative positions of oscillating systems at any given time. For pendulums, being "in phase" means that both reach their mean position simultaneously and in the same direction.Two pendulums, even if they start in phase, will eventually fall out of sync unless they complete the same fraction of their cycles in the same amount of time. In the exercise example:
  • Given the time period ratio of \(\frac{11}{10}\), the pendulums will be out of phase over successive cycles.
  • To find when they can be back in phase, we consider the number of vibrations. Here, the shorter pendulum must complete vibrations that satisfy this ratio.
By solving the equation \(\frac{n}{n+1} = \frac{10}{11}\), it determines when the shorter pendulum's vibrations bring both pendulums back in phase, specifically after 10 vibrations. This phase synchronization is essential in applications like musical instruments or engineering devices where timing is critical.

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Most popular questions from this chapter

A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is \(T\). With what acceleration should the lift be accelerated upwards in order to reduce its period to T/2? ( \(g\) is the acceleration due to gravity) [Karnataka CET 2008] (a) \(4 \mathrm{~g}\) (b) \(g\) (c) \(2 \mathrm{~g}\) (d) \(3 \bar{g}\)

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 having kinetic energy \(1 / 3\) of the maximum value at a distance of \(4 \mathrm{~cm}\) from the mean position. Find the amplitude of motion. (a) \(2 \sqrt{6} \mathrm{~cm}\) (b) \(2 / \sqrt{6} \mathrm{~cm}\) (c) \(\sqrt{2} \mathrm{~cm}\) (d) \(6 / \sqrt{2}\)

A pendulum is made to hang from a ceilling of an elevator.It has period of \(T_{\mathrm{sec}}\). (for small angles). The elevator is made to accelerate upwards with \(10 \mathrm{~m} / \mathrm{s}^{2}\).The period of the pendulum now will be (assume \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (a) \(T \sqrt{2}\) (b) infinite (c) \(T / \sqrt{2}\) (d) zero

This time period of a particle undergoing SHM is 16 s. It starts motion from the mean position. After \(2 \mathrm{~s}\), its velocity is \(0.4 \mathrm{~ms}^{-1}\). The amplitude is \(\begin{array}{ll}\text { (a) } 1.44 \mathrm{~m} & \text { (b) } 0.72 \mathrm{~m}\end{array}\) (c) \(2.88 \mathrm{~m}\) (d) \(0.36 \mathrm{~m}\)
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