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A body describes shm in a line 0.04 m lo...

A body describes shm in a line 0.04 m long. Its velocity at the centre of the line is `0.12 ms^(-1)`. Find the Time period.

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To solve the problem step by step, we will follow the given information and apply the relevant formulas for simple harmonic motion (SHM). ### Step 1: Identify the amplitude The total length of the line in which the body oscillates is given as 0.04 m. In SHM, this length represents twice the amplitude (A) of the motion. Therefore, we can calculate the amplitude as follows: \[ A = \frac{\text{Total Length}}{2} = \frac{0.04 \, \text{m}}{2} = 0.02 \, \text{m} \] ...
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