Question

The total energy of a particle executing SHM is 80 J. What is the potential energy when the particle is at a distance of 3/4 of amplitude from the mean position?

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Answer 1

Given that, the total energy is 80J.

Total energy = Kinetic energy + Potential energy

And kinetic energy is 0. So,

Total energy = 1/2 mω²A² OR 1/2 kA²

{ A = amplitude and k = positive constant }

→ 80 = 1/2 kA²

We have to find the potential energy when the particle is at a distance of 3/4 of amplitude from the mean position.

Let us assume that the particle is at a 'x' distance from the mean position.

So,

Potential energy = 1/2 kx²

And as per given condition or According to question,

x = 3/4A

So,

Potential energy = 1/2 × k × (3A/4)²

→ 1/2 × k × 9A²/16

→ 9/16 × kA²/2

→ 9/16 × 80

{ kA² = 80 }

→ 45J

Answer 2

Given :-

The total energy of a particle executing SHM is 80 J.

To find :-

The potential energy when the particle is at a distance of 3/4 of amplitude from the mean position.

Solution :-

We know,

Total energy = Kinetic energy + Potential energy

Kinetic energy = 0

[At, time T/4 , K = 1/2 m (ω)² (A² - A²) = 0]

So, total energy = potential energy

$\implies \textsf{ Total energy = }\displaystyle{\sf{\frac{1}{2}m (\omega )^2(A)^2}}$

OR

$\implies \sf{ Total\ energy =\frac{1}{2}kA^2 }$

(◙ A represents Amplitude).

$\displaystyle{\sf{\implies 80=\frac{1}{2}kA^2 }}$                       ....(i)

Let the particle be 's' distance from the mean position.

Total energy = potential energy = $\displaystyle{\sf{\frac{1}{2}kx^2}}$.

A/q, we have to find the potential energy when the particle is at a distance of 3/4 of amplitude from the mean position.

$\displaystyle{\sf{\implies x=\frac{3}{4}A}}$

So, P.E. = $\displaystyle{\sf{\frac{1}{2}k(\frac{3}{4} A)^2 }}$

$\displaystyle{\sf{\implies P.E=\frac{1}{2}k(\frac{9A^2}{16}) }}\\\\\\\displaystyle{\sf{\implies P.E=\frac{9}{16}*\frac{kA^2}{2} }}\\\\\\\displaystyle{\sf{\implies P.E=\frac{9}{16}*80 }} ..................\sf{[\frac{kA^2}{2}=80\ from\ (i)]}$

$\displaystyle{\sf{\implies P.E=9*5}}\\\\\boxed{\displaystyle{\sf{\implies P.E=45\ J}}}$