Question

The maximum velocity of a particle performing linear S.H.M. is $0.16m/s$ and its maximum acceleration is $0.64m/s^2$. The calculate its time period

• A. $1.571s$
• B. $0.08s$
• C. $16s$
• D. $3.2s$

Answer 1

The correct option is A $1.571s$

Given,

$v_{max}=0.16m/s$

$a_{max}=0.64m/s^2$

Time period, $T=?$

In S.H.M, we know that

$v_{max}=A\omega$. . . . . . .(1)

$a_{max}=A{\omega }^2$. . . . . . (2)

$\frac{a_{max}}{v_{max}}=\omega =\frac{2\pi }{T}$

$T=\frac{2\pi \times v_{max}}{a_{max}}$

$T=\frac{2\times 3.14\times 0.16}{0.64}$

$T=1.57s$

The correct option is A.