Question

A boy is executing $SHM$ uner the action of a force whose maximum magnitue is $50N$. The magnitue of force acting on the particle at the time when its half kinetic an half-potential is (Assume potential energy to be zero at mean position)

• A. $12.5\sqrt{2}N$
• B. $12.5N$
• C. $25N$
• D. $25\sqrt{2}N$

Answer 1

The correct option is D $25\sqrt{2}N$

The maximum magnitue of force on body performing SHM,
$F_{max}=m{\omega }^{2}A=50N$
According to the question at a time $t$ particle's total energy is half-kinetic energy and half-potential energy Let the particle displacement be$x$ at time $t$,
i.e.,
$PE+KE\Rightarrow \frac{1}{2}k{x}^2=\frac{1}{2}\times k(A^2-x^2)\Rightarrow x^2=\frac{A^2}{2}\Rightarrow x=\frac{A}{\sqrt{2}}$
Force can be written as,
$F=kx\Rightarrow F=m{\omega }^{2}x\Rightarrow F=m{\omega }^2\left(\frac{A}{\sqrt{2}}\right)\therefore F_{max}=m{\omega }^{2}A\Rightarrow F=\frac{F_{max}}{\sqrt{2}}\Rightarrow F=\frac{50}{\sqrt{2}}\Rightarrow F=\frac{50}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\Rightarrow F=25\sqrt{2}N$