Question

The angular velocities of three bodies in simple harmonic motion are ${\omega }_1,{\omega }_2,{\omega }_3$ with their respective amplitudes as $A_1,A_2,A_3$. If all the three bodies have same mass and velocity, the.

• A. $A_1^2{\omega }_1^2=A_2^2{\omega }_2^2=A_3^2{\omega }^2$
• B. $A_1^2{\omega }_1=A_2^2{\omega }_2=A_3^2{\omega }_3$
• C. $A_1{\omega }_1^2=A_2{\omega }_2^2=A_3{\omega }_3^2$
• D. $A_1{\omega }_1=A_2{\omega }_2=A_3{\omega }_3$

Answer 1

The correct option is D $A_1{\omega }_1=A_2{\omega }_2=A_3{\omega }_3$

Rate of change of displacement is known as velocity. The equation for displacement (y) of a body in SHM with angular velocity $\omega$ is given by
$y=a\sin \omega t$
where a is amplitude
velocity v$=$ rate of change of displacement $=\frac{dy}{dt}$
$v=\frac{dy}{dt}=\frac{d}{dt}\left(a\sin \omega t\right)=a\omega \cos \omega t$
Given, $v_1=v_2=v_3$ and $a_1=A_1,a_2=A_2$.
$a_2=A_3$
$A_1{\omega }_1=A_2{\omega }_2=A_3{\omega }_3$.