Question

The potential energy of a particle of mass 1kg in motion along the x axis is given by U =4(1-cos 2x)J , where x is in metres . The period of small oscillations (in s) is

Answer 1

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Answer 2

Answer:$T = \dfrac{\pi}{2}\ s$

Explanation:

Given that,

Potential energy U = 4(1 -cos 2x) J

Mass = 1 kg

We know that,

$F = -\dfrac{dU}{dx}$

$F = -\dfrac{d}{dx}(4-4cos2x)$

$F = -[0-4(-sin2x)2]$

$F = -8 sin(2x)$

Where, x is very small.

So, $sin(2x) = 2x$

$F = -16 x\ N$

Now,  $F = ma$

$-16x\ N = 1kg\times a$

$a = -16 x\ m/s^{2}$

The acceleration of the simple harmonic oscillation

$a =-\omega^{2}x$

$-16x = \omega^{2}x$

$\omega^{2} = 16$

$\omega = 4$

The time period of small oscillations is

$\omega = \dfrac{2\pi}{T}$

$T = \dfrac{\pi}{2}\ s$

Hence, the time period of small oscillations is $\dfrac{\pi}{2}\ s$.