Answer to Question #89774 in Mechanics | Relativity for Shivam Nishad

Question

Simple harmonic motion is given by
x = 3 sin 4πt + 4 cos 4πt where x is in metre
and t in second. Calculate
(i) amplitude of oscillator,
(ii) frequency of oscillation and
(iii) displacement at time t =0

Answer 1

The position of SHO is given by equation

$$x(t)=3\sin (4\pi t)+4\cos(4\pi t)$$

We put

$$x(t)=A(\cos\phi\sin (4\pi t)+\sin\phi\cos(4\pi t))$$ $$=A\sin(4\pi t+\phi)$$

Here

$$A\cos\phi=3,\quad A\sin\phi=4$$

(i) So, the amplitude of oscillation

$$A=\sqrt{3^2+4^2}=5\:\rm{m}$$

The initial phase

$$\phi=\tan^{-1}\frac{4}{3}$$

Therefore the equation of motion of SHO

$$x(t)=A\sin(\omega t+\phi)=5\sin(4\pi t+\phi)$$

(ii) The frequency of oscillation

$$f=\frac{\omega}{2\pi}=\frac{4\pi}{2\pi}=2\:\rm{Hz}$$

(iii) The displacement at time t=0

$$x(0)=3\sin (4\pi \times 0)+4\cos(4\pi \times 0)=4\:\rm{m}$$