Question

A point particle of mass $0.1kg$ is executing SHM of amplitude $0.1m$. When the particle passes through the mean position, its KE is $8\times {10}^{-3}J$. The equation of motion of this particle phase of oscillation is ${45}^o$ is

• A. $y=0.1\sin (\frac{t}{4}+\frac{\pi }{4})$
• B. $y=0.1\sin (\frac{t}{2}+\frac{\pi }{4})$
• C. $y=0.1\sin (4t-\frac{\pi }{4})$
• D. $y=0.1\sin (4t+\frac{\pi }{4})$

Answer 1

The correct option is D $y=0.1\sin (4t+\frac{\pi }{4})$

KE at mean position is given by:

$KE=\frac{1}{2}m{\omega }^2{a}^2=8\times {10}^{-3}$

or, $\omega =\left(\frac{2\times 8\times {10}^{-3}}{m{a}^2}\right)=[\frac{2\times 8\times {10}^{-3}}{0.1\times \left({0.1}^2\right)}{]}^{\frac{1}{2}}=4$

So, equation of SHM is:

$y=a\sin (\omega t+\theta )=0.1\sin (4t+\frac{\pi }{4})$