Question

The maximum velocity of a particle performing s.h.m. is 6.28cm/s. If the length of its path is 8 cm, calculate its period

Answer 1

Given :

Maximum velocity of a particle performing SHM is 6.28cm/s.

Path length is 8 cm.

To find :

The time period of the particle

Solution :

We know velocity in simple harmonic motion is given as,

v = Aω

here,

• v denotes velocity in SHM.
• A denotes amplitude.
• ω denotes angular frequency.

we know,

Amplitude, A = path length/2 = 8/2 = 4cm.

Angular frequency, ω = 2π/T (here, T denotes time period)

Now, substituting all the given values in the formula,

$: \implies{ \sf{v = A \omega}}$

$: \implies{ \sf{6.28 = 4 \times \dfrac{2\pi}{T} }}$

$: \implies{ \sf{ \dfrac{6.28}{4} = \dfrac{2\pi}{T} }}$

$: \implies{ \sf{1.57 = \dfrac{2\pi}{T} }}$

$: \implies{ \sf{T = \dfrac{2\pi}{1.57} }}$

$: \implies{ \sf{T = \dfrac{2 \times 3.14}{1.57} }}$

$: \implies \underline{ \boxed{ \sf{ \gray {T = 4s}}}} \dag$