Question

A particle executes SHM with a period 6 sec. and amplitude 0.03m.
speed in cm/sec is
(2) a
(1) 1/2
(4) 3 1/2
(3) 20
​

Answer 1

HEYA GM ❤

★ T = 2π / Ω

Ω = 2π /T = 2π / 6 = π /3

★ VELOCITY = AMPLITUDE × ANGULAR FREQUENCY

V = 0.03m × π/3

V = 3cm × π/3 sec

V = π = 3.14 cm / s

TQ ❤ ❤

Answer 2

The speed of the particle in SHM is 3.14 cm/s.

Explanation:

It is given that,

Period of particle in SHM, T = 6 s

Amplitude of particle, A = 0.03 m

Let v is the speed of particle in SHM. It is given by the below relation as :

$v=A\times \omega$

$v=A\times \dfrac{2\pi}{T}$

$v=0.03\times \dfrac{2\pi}{6}$

v = 0.0314 m/s

or

v = 3.14 cm/s

So, the speed of the particle in SHM is 3.14 cm/s. Hence, this is the required solution.

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Simple harmonic motion

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