A particle is executing a simple harmonic motion. Its maximum acceleration is $α$ and maximum velocity is $β$ . Then, its time period of vibration will be:
$\frac{β^{2}}{α^{2}}$
$\frac{2 π β}{α}$
$\frac{β^{2}}{α}$
$\frac{α}{β}$

Question

A particle is executing a simple harmonic motion. Its maximum acceleration is $α$ and maximum velocity is $β$ . Then, its time period of vibration will be:
$\frac{β^{2}}{α^{2}}$
$\frac{2 π β}{α}$
$\frac{β^{2}}{α}$
$\frac{α}{β}$

Toppr Admin · Accepted Answer

For SHM- maximum acceleration-xA0-amax-x3C9-2a-x3B1- -xA0-xA0-where a is the amplitude of SHM-and-maximum velocity- vmax-x3C9-a-x3B2-so- -x3B1-x3B2-x3C9-2a-x3C9-a-x3C9-or -x3B1-x3B2-2-x3C0-T-or T-2-x3C0-x3B2-x3B1-

A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β. Then, its time period of vibration will be:β2α22πβαβ2ααβ

For SHM, maximum acceleration, amax=ω2a=α where a is the amplitude of SHM. and maximum velocity, vmax=ωa=β so, αβ=ω2aωa=ω or αβ=2πT or T=2πβα

A particle is executing a simple harmonic motion. Its maximum acceleration is $α$ and maximum velocity is $β$ . Then, its time period of vibration will be:
$\frac{β^{2}}{α^{2}}$
$\frac{2 π β}{α}$
$\frac{β^{2}}{α}$
$\frac{α}{β}$

For SHM, maximum acceleration, $a_{m a x} = ω^{2} a = α$ where a is the amplitude of SHM.
and maximum velocity, $v_{m a x} = ω a = β$
so, $\frac{α}{β} = \frac{ω^{2} a}{ω a} = ω$
or $\frac{α}{β} = \frac{2 π}{T}$
or $T = \frac{2 π β}{α}$