First slide

Simple harmonic motion

Question

A particle of mass 0.8 kg is executing simple harmonic motion with an amplitude of 1.0 metre and periodic time 11/7 sec. The kinetic energy (in J) of the particle at the moment when it displacement is 0.6 metre is ________.

Moderate

Solution

We know that at a displacement y from mean position particle's velocity is given as
$v = ω \sqrt{(A^{2} - y^{2})}$
$or v = \frac{2 π}{T} \sqrt{(A^{2} - y^{2})}$
$= \frac{2 \times 3 .14}{(11 / 7)} \sqrt{[(1 .0)^{2} - (0 .6)^{2}]} = 3 .2 m / s$
Kinetic energy at this displacement is given by
$K = \frac{1}{2} {mv}^{2} = \frac{1}{2} \times 0 .8 \times (3 .2)^{2} = 4 .1 J$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App