Question

A body vibrates In S. H. M with a frequency of 50hz and an amplitude of 4cm find the period, acceleration at the middle and at the end of the path of oscillation

Answer 1

How to solve it

Answer 2

Solution

Answer 3

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Answer 4

[AI-generated answer — Explain Bot]

To find the period of the body's oscillation, we can use the formula:

Period (T) = 1 / Frequency (f)

Given that the frequency is 50 Hz, we can substitute this value into the formula:

T = 1 / 50

Simplifying the expression:

T = 0.02 seconds

So, the period of the body's oscillation is 0.02 seconds.

To find the acceleration at the middle of the path of oscillation, we need to recall that in simple harmonic motion (SHM), the acceleration is given by the formula:

Acceleration (a) = (2πf)^2 * amplitude

Using this formula, we can substitute the values:

Amplitude = 4 cm

Frequency = 50 Hz

π ≈ 3.14

a = (2π * 50)^2 * 4

Simplifying the expression:

a = (100π)^2 * 4

a = 10000π^2 cm/s^2

So, the acceleration at the middle of the path of oscillation is 10000π^2 cm/s^2.

To find the acceleration at the end of the path of oscillation, we need to note that at the extreme points of the motion, the acceleration is equal to zero. Therefore, the acceleration at the end of the path of oscillation is zero.