Question 157: Oscillations

Question

The velocity of a particle executing simple harmonic motion along $x$-axis is described as $v^2=50-x^2$, where $x$ represents displacement. If the time period of motion is $\frac{x}{7} \mathrm{~s}$, the value of $x$ is $\_\_\_\_$ .

Accepted Answer

For a particle in simple harmonic motion, the velocity is given by
$$ v^2=\omega^2\left(A^2-x^2ight) $$
We are given
$$ v^2=50-x^2 $$
Comparing this with the standard SHM form,
$$ v^2=\omega^2(A^2-x^2) $$
we get
$$ \omega^2=1 $$
So,
$$ \omega=1\ 	ext{rad/s} $$
Now, time period of SHM is
$$ T=\frac{2\pi}{\omega} $$
Therefore,
$$ T=\frac{2\pi}{1}=2\pi\ 	ext{s} $$
Given in the question that time period is
$$ T=\frac{x}{7}\ 	ext{s} $$
So,
$$ \frac{x}{7}=2\pi $$
Hence,
$$ x=14\pi $$
Using $\pi=\frac{22}{7}$,
$$ x=14	imes \frac{22}{7}=44 $$
So, the required value is
$$ \boxed{44} $$

Oscillations

The velocity of a particle executing simple harmonic motion along x-axis is described as v^2=50-x^2, where x represents displacement. If the time period of motion is \frac{x}{7} \mathrm{~s}, the value of x is \_\_\_\_ .

Integer / Numeric Answer Type