Question

A certain simple harmonic vibrator of mass $0.1kg$ has a total energy of $10J$.Its displacement from the mean position is $1cm$ when it has equal kinetic and potential energies. The amplitude $A$ and frequency $n$ of vibration of the vibrator are

• A. $A=\sqrt{2}cm,$ $n=\frac{500}{\pi }Hz$
• B. $A=\sqrt{2}cm,n=\frac{1000}{\pi }Hz$
• C. $A=\frac{1}{\sqrt{2}}cm,n=\frac{500}{\pi }Hz$
• D. $A=\frac{1}{\sqrt{2}}cm,n=\frac{1000}{\pi }Hz$

Answer 1

The correct option is A $A=\sqrt{2}cm,$ $n=\frac{500}{\pi }Hz$

For a displacement $x$, the kinetic and potential energies are $KE=\frac{1}{2}m(A^2-x^2){\omega }^2$ and $PE=\frac{1}{2}m{x}^2{\omega }_2^2$

Each of these $=10/2=5J$ when $x=1cm$

Hence $\frac{1}{2}m{x}^2{\omega }^2=\frac{1}{2}\times 0.1\times (1\times {10}^{-2}{)}^2{\omega }^2=5$

This gives ${\omega }^2=\frac{10}{0.1\times {10}^{-4}}={10}^6$
Giving ${\omega }_1={10}^3=1000rad/s$

Hence $T=\frac{2\pi }{\omega }=\frac{\pi }{500}s$
And frequency $f=\frac{500}{\pi }s$

Also, $KE=5=\frac{1}{2}\times 0.1[A^2-(1\times {10}^{-2}{)}^2\left]\right({10}^6)$

$A^2-{10}^{-4}={10}^{-4}$

$A^2=2\times {10}^{-4}$

$A^2=2\times {10}^{-4}\Rightarrow A=\sqrt{2}\times {10}^{-2}m=\sqrt{2}cm$