Question

A particle is executing S.H.M. of amplitude 5cm and period of 2sec. find the speed of the particle at a point where it's acceleration is half of its maximum value​

Answer 1

The speed of the particle at a point is 13.6 cm/s.

Explanation:

Given that,

Amplitude = 5 cm

Time period = 2 sec

The acceleration is half of its maximum value.

$x=\dfrac{A}{2}$

We need to calculate the speed of the particle at a point

Using formula of velocity

$v=\omega\sqrt{(A)^2-(x)^2}$

$v=\omega\sqrt{(A)^2-(\dfrac{A}{2})^2}$

Put the value into the formula

$v=\dfrac{2\pi}{2}\sqrt{(5)^2-(\dfrac{5}{2})^2}$

$v=13.6\ cm/s$

Hence, The speed of the particle at a point is 13.6 cm/s.

Learn more :

Topic : speed

https://brainly.in/question/6568026