Question

The maximum velocity of a particle executing simple harmonic motion is $v$. If the amplitude is doubled and the time period of oscillation decreased to $1/3$ of its original value, the maximum velocity becomes:

• A. $18v$
• B. $12v$
• C. $6v$
• D. $3v$

Answer 1

The correct option is C $6v$

Lets consider $v=$the maximum velocity of a particle executing harmonic motion

$v=A\omega$. . . . . . . .(1)

After the time decreased $1/3$ of its original value,

$T^{\prime }=\frac{1}{3}T$

$A^{\prime }=2A$

${\omega }^{\prime }=\frac{2\pi }{T^{\prime }}=3\omega$

$v^{\prime }=A^{\prime }{\omega }^{\prime }$

$v^{\prime }=2A\times 3\omega$

$v^{\prime }=6A\omega =6v$ (from equation 1)

The correct option is C.