Question:

Two pendulums of lengths 100 cm and 121 cm start vibrating. At some instant the two are at the mean position in the same phase. After how many vibrations of the longer pendulum will the two be in the same phase at the mean position again?

Updated On: Jun 23, 2023
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The Correct Option is A

Solution and Explanation

Let the two pendulums are in same phase, after n vibrations of the longer pendulum. In this time the shorter pendulum will complete $ (n+1) $ vibrations. $ \therefore $ $ n2\pi \sqrt{\frac{{{l}_{2}}}{g}}=(n+1)2\pi \sqrt{\frac{{{l}_{1}}}{g}} $ $ n2\pi \sqrt{\frac{121}{g}}=(n+1)2\pi \sqrt{\frac{100}{g}} $ $ 11n=10(n+1) $ $ n=10 $
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Concepts Used:

Oscillations

Oscillation is a process of repeating variations of any quantity or measure from its equilibrium value in time . Another definition of oscillation is a periodic variation of a matter between two values or about its central value.

The term vibration is used to describe the mechanical oscillations of an object. However, oscillations also occur in dynamic systems or more accurately in every field of science. Even our heartbeats also creates oscillations​. Meanwhile, objects that move to and fro from its equilibrium position are known as oscillators.

Read More: Simple Harmonic Motion

Oscillation- Examples

The tides in the sea and the movement of a simple pendulum of the clock are some of the most common examples of oscillations. Some of examples of oscillations are vibrations caused by the guitar strings or the other instruments having strings are also and etc. The movements caused by oscillations are known as oscillating movements. For example, oscillating movements in a sine wave or a spring when it moves up and down. 

The maximum distance covered while taking oscillations is known as the amplitude. The time taken to complete one cycle is known as the time period of the oscillation. The number of oscillating cycles completed in one second is referred to as the frequency which is the reciprocal of the time period.