Question

A particle of mass 1 kg is performing SHM and its displacement any time is given by

What will be kinetic energy of particle at t = 2 sec?​

Answer 1

SOLUTION

GIVEN

A particle of mass 1 kg is performing SHM and its displacement any time is given by

$\displaystyle \sf{y = 2 \sin \bigg( \: \frac{\pi t}{12} + \frac{\pi}{6} \: \bigg)}$

TO DETERMINE

The kinetic energy of particle at t = 2 sec

EVALUATION

Here the given equation is

$\displaystyle \sf{y = 2 \sin \bigg( \: \frac{\pi t}{12} + \frac{\pi}{6} \: \bigg)}$

Comparing with the general equation

$\displaystyle \sf{y = a \sin ( \omega t + \phi)}$

We get

$\displaystyle \sf{a = 2 \: ,\: \: \omega = \frac{\pi }{12} \: ,\: \phi = \frac{\pi}{6} \:}$

Now for t = 2 we have

$\displaystyle \sf{y = 2 \sin \bigg( \: \frac{2\pi }{12} + \frac{\pi}{6} \: \bigg)}$

$\displaystyle \sf{ \implies \: y = 2 \sin \bigg( \: \frac{\pi }{6} + \frac{\pi}{6} \: \bigg)}$

$\displaystyle \sf{ \implies \: y = 2 \sin \bigg( \: \frac{\pi }{3} \: \bigg)}$

$\displaystyle \sf{ \implies \: y = 2 \times \frac{ \sqrt{3} }{2} }$

$\displaystyle \sf{ \implies \: y = \sqrt{3} }$

Hence the required kinetic energy

$\displaystyle \sf{ = \frac{1}{2}m { \omega}^{2} ( {a}^{2} - {y}^{2} ) \: \: J }$

$\displaystyle \sf{ = \frac{1}{2} \times 1 \times { \bigg( \frac{\pi}{12} \bigg)}^{2} \times \bigg[ {2}^{2} - {( \sqrt{3}) }^{2} \bigg] \: \: J }$

$\displaystyle \sf{ = \frac{1}{2} \times \frac{ {\pi}^{2} }{144} \times \bigg[ 4 - 3 \bigg] \: \: J }$

$\displaystyle \sf{ = \frac{ {\pi}^{2} }{288} \: \: J }$

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. find the speed of sound in air at 50 degree C and 70 dgree C (take speed of sound 332m/s)

https://brainly.in/question/40606714

2. A simple pendulum has period 4 s. The length. of the string of the pendulum will be (take g = 10 m/s2 and 1T2 = 10)

https://brainly.in/question/36767516