Question

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

$E=\frac{1}{2}m{\omega }^2{A}^2\Rightarrow E=\frac{1}{2}m(2\pi f{)}^2{A}^2\Rightarrow A=\frac{1}{2\pi f}\sqrt{\frac{2E}{m}}PuttingE=K+Uweobtain,A=\frac{1}{2\pi \left(\frac{25}{\pi }\right)}\sqrt{\frac{2\times (0.5+0.4)}{0.2}}\Rightarrow A=0.06m$