Question

An object of mass 0.2 kg executes SHM along x-axis with a frequency of $\frac{25}{\pi }$ Hz. At the position x=0.04 m the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation will be

• A. 0.06 m
• B. 0.04 m
• C. 0.05 m
• D. 0.25 m

Answer 1

The correct option is A 0.06 m

At $x=0.01m$, the Total energy $=K.E+P.E$

$T.E=0.5J+0.4J=0.9J$

we know that $T.E$ of $SHM=\frac{1}{2}m{w}^2{A}^2$

$m=0.2,w=2\pi f=50$

$\Rightarrow \frac{1}{2}m{w}^2{A}^2=0.9$

$\frac{1}{2}\times 0.2\times (50{)}^2{A}^2=0.9$

$A=0.06m\left(A\right)$