Question

Two particles are executing SHM, one with angular frequency $\omega$ and the other with ${}^{\prime }2{\omega }^{\prime }$, which of the following $v-x$ graph correctly represent simple hormonic motions of two particles of same amplitude.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

We know that,
$v^2={\omega }^2(A^2-x^2)$
For particle $1;v_1^2={\omega }^2(A^2-x^2)...\left(1\right)$
For particle $2;v_2^2=4{\omega }^2(A^2-x^2)...\left(2\right)$
From (1) and (2) we can infer,
at, $x=0;v_1=\pm A\omega ;v_2=\pm 2\omega A$
at, $x=\pm A;v_1=0;v_2=0$
Hence, option (c) is correct