Question

A partical perform SHM with a period of 8 s. if the particle starts from the mean position at what time during one oscillation is its energy half potential?

Answer 1

Answer:

1 s

Explanation:

T = 8s

w=2π/8=π/4

PE = 1/2 PE

1/2 mw2x2= 1/2(1/2 mw2a2)

x2=a2/2

x=a/√2

x=a sin wt

a/√2= a sin π/4 t

1/√2=1/√2 t

t=1 s

Answer 2

1 s.

Explanation:

The kinetic energy in S.H.M is given as

$K.E=\frac{1}{2}k(A^2-y^2)$

Here, y is distance from mean position and A is the amplitude.

The potential energy in S.H.M is given as

$P.E=\frac{1}{2}ky^2$

As per question,

P.E = K.E

$\frac{1}{2}ky^2 =\frac{1}{2}k(A^2-y^2)$

$y=\frac{A}{\sqrt{2} }$$\frac{A}{\sqrt{2} } =A\ sin(\frac{2\pi}{T})t$

General equation,

$y=A sin\omega t$

$y=Asin(\frac{2\pi}{T})t$

Given, T = 8 s.

So,

$\frac{A}{\sqrt{2} } =Asin(\frac{2\pi}{8s})t$

or,   $sin(\frac{\pi}{4} }) =sin(\frac{2\pi}{8s})t$

or, $\frac{\pi}{4}=\frac{\pi t}{4}$

t = 1 s.

Thus, at 1 s during one oscillation is its energy half potential.

#Learn More: Energy in SHM.

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