Question

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

y=2sin(pi/12 t + pi/6)m

What will be kinetic energy of particle at t = 2 sec?
1.pi^2/288J
2.pi^/144J
3.pi/288J
4.pi/144J

Answer 1

Answer:

pi^2 / 288

Explanation:

differentiate the given y eqn

v= dy/dt----------> 1

K=1/2 mv^2---------> 2

Sub eqn 1 in eqn 2,

We get answer as pi^2/288

Answer 2

Answer:

The kinetic energy of the particle at t = 2 sec will be $\frac{\pi^2}{96}$ i.e.none of the options are correct.

Explanation:

The kinetic energy of the body at a particular instant is given as,

$K=\frac{1}{2}m\omega^2y^2$       (1)

Where,

K=jinetic energy of the body

m=mass of the body

ω=angular frequency of the body

y=instantaneous displacement of the body

The waveform given in the question is,

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

By substituting the value of "t" i.e. 2 seconds in equation (2) we get;

$y=2sin(\frac{\pi\times 2}{12}+\frac{\pi}{6})$

$y=2sin(\frac{\pi}{3})$

$y=2\times sin60\textdegree$                ($sin60\textdegree=\frac{\sqrt{3} }{2}$)

$y=\sqrt{3}$       (3)

The equation (2) is of the form,

$y=Asin(\omega t+\phi)$       (4)

So, by comparing equation (2) with equation (3) we can say,

A=2m

$\omega=\frac{\pi}{12}$

$\phi=\frac{\pi}{6}$

By substituting all the required values in equation (1) we get;

$K=\frac{1}{2}\times 1\times (\frac{\pi}{12} )^2\times (\sqrt{3}) ^2$

$K=\frac{\pi^2}{288} \times 3$

$K=\frac{\pi^2}{96}$

Hence, the kinetic energy of the particle at t = 2 sec will be $\frac{\pi^2}{96}$ i.e.none of the options are correct.