Question

A harmonic oscillator vibrates with amplitude of 4 cm and performs 150 oscillations in one minute. If the initial phase is $45\circ$ and it starts moving away from the equation of motion is

• A. $0.04sin(5\pi t+\frac{\pi }{4})$
• B. $0.04sin(5\pi t-\frac{\pi }{4})$
• C. $0.04sin(4\pi t+\frac{\pi }{4})$
• D. $0.04sin(4\pi t-\frac{\pi }{4})$

Answer 1

The correct option is A $0.04sin(5\pi t+\frac{\pi }{4})$

Equation of motion of harmonic oscillator is $y=Asin(\omega t+{\varphi }_0)$

It is given that $A=0.04m$

and ${\varphi }_0={54}^{\circ }=\frac{45}{180}\pi =\frac{\pi }{4}$

$\omega =2\pi \nu =2\pi \times \frac{150}{60}=5\pi$

Thus $y=0.04sin(5\pi t+\frac{\pi }{4})$

Hence the answer is option A.