Question

amplitude of oscillation of a particle that executes SHM is 2 cm its displacement from its mean position in a time equal to 1/6th of its time period is

Answer 1

Answer:

root 3

Explanation:

x=A sin (wt)

x=2 sin(w*T/6)

T=2 pi/w

x=2 sin(pi/3)

x=2*root 3/2

x=root 3 cm

Answer 2

Answer:

The displacement of the particle from its mean position is $\sqrt{3}$.

Explanation:

It is given that,

Amplitude of oscillation, A = 2 cm = 0.02 m

The displacement of the particle is given by :

$y=A\ sin\omega t$

$\omega$ is the angular frequency

Here, $t=\dfrac{T}{6}$

$y=2\ sin(\dfrac{2\pi}{T}\times \dfrac{T}{6})$

$y=2\ sin(\dfrac{\pi}{3})$

$y=2\times \dfrac{\sqrt{3} }{2}$

$y=\sqrt{3}$

So, the displacement of the particle from its mean position is $\sqrt{3}$. Hence, this is the required solution.

Learn more,

Simple harmonic motion

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