Question

A point particle of mass 0.1kg is executing SHM with amplitude of 0.1m.When the particle passes through the mean position, its K.E is $8\times {10}^{-3}J$.Obtain the equation of motion of this particle if the initial phase of oscillation is ${45}^{\circ }$.

• A. $y=0.2sin(4t+\pi /2)$
• B. $y=0.1sin(4t+\pi /4)$
• C. $y=0.2sin(4t+\pi /4)$
• D. None of these

Answer 1

The correct option is B $y=0.1sin(4t+\pi /4)$

KE at mean is maximum.
Thus, $\frac{1}{2}m{v}_{max}^2=8\times {10}^{-3}$ or $v_{max}=\sqrt{16\times {10}^{-2}}=0.4m/s$
WE know that $x=Asin(\omega t+\theta )$
Thus $v=dx/dt=A\omega cos(\omega t+\theta )$, which has the maximum value when cosine is 1, i.e $A\omega$
Thus, $A\omega =0.4$
Given A=0.1, thus $\omega =4$
Initial phase is ${45}^0=\pi /4$
Thus B is the correct option.