Question

The potential energy of a simple harmonic oscillator of mass 2 kg at its mean position is 5j.If its total energy is 9j and its amplitude is 0.01m, its time period would be:

Answer 1

Answer : Time period = π/100 s

Solution :
_________

Given that : The potential energy of a simple harmonic oscillator of mass 2 kg at its mean position is 5j

K.E at the centre= 9-5 = 4J

According to the question :

$\frac{1}{2} k {A}^{2} = (9 - 5)$

where, k = kinetic energy and A = amplitude

$= > k = \frac{8}{ {A}^{2} } \\ \\ = > k = \frac{8}{ {(0.01)}^{2} } \\ \\ = > k = 8 \times {10}^{4}$

=> k = 8×10⁴ N/m

As we know that :

$t = 2\pi \sqrt{ \frac{m}{k} }$

Where, T = Time period

$= > T = 2\pi \sqrt{ \frac{2}{8 \times {10}^{4} } } \\ \\ = > T = \frac{\pi}{100} s$

So, the time period would be π/100 s

Answer 2

answer : π/100 sec

explanation : amplitude, A = 0.01m

mass of oscillator , m = 2 kg

potential energy at mean position, P.E = 5J

total energy at mean position , T.E = 9J

at mean position,

total energy = potential energy + kinetic energy

9J = 5J + kinetic energy

kinetic energy = 4J

or, 1/2 mv² = 4J

or, 1/2 × 2 × v² = 4

or, v² = 4

hence, v = 2m/s , it is maximum speed of oscillator.

so, $v=\omega A$

2 = $\omega$ × 0.01

or, 200 = $\omega$

we know, $T=\frac{2\pi}{\omega}$

so, $T=\frac{2\pi}{200}=\frac{\pi}{100}$

hence, time period is π/100 sec