Question

The maximum velocity of a particle
performing S.H.M.is 6.28 cm/s. If the length
of its path is 8 cm, calculate its period.​

Answer 1

v= 6.28

l of path = 8 cm

that means Amplitude A = 4 cm

v=omega*A

omega= v/A =6.28/4

time period= 2π/(omega)

= 6.28/(omega) =4 s

Answer 2

The time period of oscillation is 4 s.

Explanation:

The maximum velocity of a particle  executing S.H.M  is given as

$v_{max}=\omega A$

Here,  A is the amplitude of S.H.M and $\omega$ is the angular frequency.

As path length is 8 cm, so

amplitude, A = 8 cm/2 =4 cm.

Given $v_{max} =6.28cm/s$.

substitute the value, we get

$6.68cm/s=\frac{2\pi}{T}\times4cm$                            (As  $\omega=\frac{2\pi}{T}$, T is the time period)

$T=\frac{2\times3.14}{6.68cm/s}\times4cm$

T = 3.76 s = 4 s.

Thus, time period of oscillation is 4 s.

#Learn More: maximum velocity in S.H.M.

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