Question

write the equation of motion of a simple harmonic oscillator which has a amplitude of 5cm and it executes 150 oscillations in 5minutes with an initial phase of 45 degree.also obtain the value of its maximum velocity.

Answer 1

Question 33949 General equation of harmonic motion is $x = A\cos (\omega t + \delta)$. Cyclic frequency might be found knowing the period $T$: $\omega = \frac{2\pi}{T}$. Since there were 150 oscillations in $5\mathrm{min}$, period is $T = 2s$ and frequency is $\omega = \frac{2\pi}{4} = \pi$. The amplitude is $A = 5$ and initial phase is $\delta = \frac{\pi}{4}$. Hence, the equation of motion is $x = 5\cos (\pi t + \frac{\pi}{4})$. Maximum velocity is $v_{max} = \omega A = 5\pi$.