Question

A point particle of mass 0.1Kg is executing SHM of amplitude 0.1m. when the particle passes through the mean position, Its kinetic energy is $8 \times {10^{ - 3}}J$. The equation of motion of this particle phase of oscillation is ${45^ \circ }$is-
A. $y = 0.1\sin \left( {\dfrac{t}{4} + \dfrac{\pi }{4}} \right)$
B. $y = 0.1\sin \left( {\dfrac{t}{2} + \dfrac{\pi }{4}} \right)$
C. $y = 0.1\sin \left( {4t - \dfrac{\pi }{4}} \right)$
D. $y = 0.1\sin \left( {4t + \dfrac{\pi }{4}} \right)$

Answer 1

Hint: Use the equation of the displacement of a wave and calculate the velocity. Then compare the kinetic energy of the wave with given value to calculate the angular speed.

Complete step-by-step answer:
Step1: Use the displace equation of a particle executing SHM-
$y = a.\sin (\omega t - \phi )$ ……..(1)
Where a= amplitude, $\phi$= phase difference=${45^ \circ }$
Now differentiate with respect to t to calculate the velocity,
$velocity = \dfrac{{dy}}{{dt}} = \omega a.\cos \left( {\omega t - \phi } \right)$
Now the maximum value of velocity is given by-
${v_{\max }} = \omega a$
Step2: Now from the equation of kinetic energy calculate the $\omega$-
$\dfrac{1}{2}mv_{\max }^2 = 8 \times {10^{ - 3}}$
Substitute the values of m and v in above equation we get,
$\dfrac{1}{2} \times 0.1 \times {\omega ^2}{a^2} = 8 \times {10^{ - 3}} \\\ \dfrac{1}{2} \times 0.1 \times {\omega ^2} \times {(0.1)^2} = 8 \times {10^{ - 3}} \\\ {\omega ^2} = 16 \\\ \Rightarrow \omega = \sqrt {16} \\\ \omega = 4 \\\$
Step3: Now substitute all the values in equation(1) to calculate the equation of the wave.
Therefore,
$y = 0.1\sin \left( {4t + \dfrac{\pi }{4}} \right)$
Which is the required equation, hence option (D) is the correct option.

Note: Keep in mind that while writing the equation always convert the phase angle in radians not in degrees.