A particle executes SHM with the time period of 2 s and amplitude 5cm. Maximum magnitude of its velocity is
$20 π c m s^{- 1}$
$2.5 π c m s^{- 1}$
$5 π c m s^{- 1}$
$10 π c m s^{- 1}$

Question

A particle executes SHM with the time period of 2 s and amplitude 5cm- Maximum magnitude of its velocity is 20pi cm s-1-2-5pi cm s-1-5pi cm s-1-10pi cm s-1-

Toppr Admin · Accepted Answer

Vmax-A-x3C9-x3C9-2-x3C0-2-x3C0-s-x2212-1A-5cm-x21D2-Vmax-5cm-xD7-x3C0-s-x2212-1-5-x3C0-cms-x2212-1

A particle executes SHM with the time period of 2 s and amplitude 5cm. Maximum magnitude of its velocity is 20πcms−12.5πcms−15πcms−110πcms−1

Vmax=Aωω=2π2=πs−1A=5cm⇒Vmax=5cm×πs−1=5πcms−1

A particle executes SHM with the time period of 2 s and amplitude 5cm. Maximum magnitude of its velocity is
$20 π c m s^{- 1}$
$2.5 π c m s^{- 1}$
$5 π c m s^{- 1}$
$10 π c m s^{- 1}$

$V_{m a x} = A ω$
$ω = \frac{2 π}{2} = π s^{- 1}$
$A = 5 c m$
$\Rightarrow V_{m a x} = 5 c m \times π s^{- 1} = 5 π c m s^{- 1}$