Question

An object of mass $0.2kg$ executes simple harmonic motion along the X-axis with a frequency of $\frac{25}{\pi }Hz.$ At the position, $x=0.04m,$ the object has a kinetic energy of $0.5J$ and potential energy of $0.4J,$ the amplitude of oscillation in the meter is equal to :

• A. $0.05$
• B. $0.06$
• C. $0.01$
• D. None of these

Answer 1

The correct option is B $0.06$

$E=\frac{1}{2}m{\omega }^2{A}^2$

$\Rightarrow E=\frac{1}{2}m(2\pi f{)}^2{A}^2$

$\Rightarrow A=\frac{1}{2\pi f}\sqrt{\frac{2E}{m}}$

Putting $E=K+U$ we obtain,

$A=\frac{1}{2\pi \left(\frac{25}{\pi }\right)}\sqrt{\frac{2\times (0.5+0.4)}{0.2}}$

$\Rightarrow A=0.06m$