Question 176: Oscillations

Question

The general displacement of a simple harmonic oscillator is $$x = A\sin \omega t$$. Let T be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when $$t = {T \over \beta }$$. The value of $$\beta$$ is ______________.

Accepted Answer

$$U = {1 \over 2}m{\omega ^2}{A^2}{\sin ^2}\omega t$$
So, $${{dU} \over {dt}} = {{m{\omega ^3}{A^2}} \over 2}\sin 2\omega t$$
This value will be maximum when $$\sin 2\omega t = 1$$
or $$2\omega t = {\pi  \over 2}$$
$$2 	imes {{2\pi } \over T}t = {\pi  \over 2}$$
$$ \Rightarrow t = {T \over 8}$$
So $$\beta  = 8$$

Oscillations

The general displacement of a simple harmonic oscillator is $x = A\sin \omega t. Let T be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when t = {T \over \beta }. The value of \beta$ is ______________.

Integer / Numeric Answer Type