Question

MCG (100)
Q17. A particle executing SHM has time
period of 8 sec and amplitude 0.40
metre. Its maximum acceleration is
(A) 0.4268 ms-2
(B) 0.2468 ms-2
(C) 0.6248 ms-2
(D) 0.8642 ms-2​

Answer 1

Given:

A particle executing SHM has time period of 8 sec and amplitude 0.40 metre.

To find:

Maximum acceleration ?

Calculation:

First of all, let's calculate the angular frequency of the particle performing SHM :

$\sf \therefore \: T = \dfrac{2\pi}{ \omega}$

$\sf \implies \: 8 = \dfrac{2\pi}{ \omega}$

$\sf \implies \: \omega = \dfrac{\pi}{4} \:hz$

Now , max acceleration be a :

$\sf \therefore \: | a| = { \omega}^{2} A$

$\sf \implies\: | a| = {( \dfrac{\pi}{4}) }^{2} \times 0.4$

$\sf \implies\: | a| = 0.616 \times 0.4$

$\sf \implies\: | a| = 0.2468 \: m {s}^{ - 2}$

So, max acceleration of SHM is 0.2468 m/s².

Answer 2

The maximum acceleration is a = 0.2468 m/s^2

Explanation:

We are given that:

The time period of the particle executing simple harmonic motion = 8 seconds

The amplitude of the particle = 0.40 m

Solution:

The formula to find the maximum acceleration is

T = 2 π / ω

8 = 2 π / ω

Now

ω = π / 4 Hz

a = ω^2 A

a = (π / 4)^2 x 0.4

a = 0.616 x 0.4

a = 0.2468 m/s^2