Question

A point mass 0.1 kg in SHM of amplitude 0.1 m. When the particle passes through the mean position, its kinetic energy is 8x10${}^{-3}$J. The equation of motion of this particle if the initial phase of oscillation is ${45}^{\circ }$ is:

• A. y=0.1 sin (4t + $\pi$ /4)
• B. y=0.1cos (4t + $\pi$/4)
• C. y=0.1 sin (2t + $\pi$ /4
• D. y=0.1cos (2t + $\pi$ /4)

Answer 1

Answer: (A)

y=0.1 sin (4t + $\pi$ /4)

$KE=\frac{1}{2}m{\omega }^2(A^2-x^2)$
x=0
$8\times {10}^{-3}=\frac{1}{2}\times 0.1\times {\omega }^2\times (0.1{)}^2$
$\omega =4$
$y=Asin\left(\omega t{+}^{\pi }{/}_4\right)$
$\therefore y=0.1sin\left(4t{+}^{\pi }{/}_4\right)$