Question

SHM of same amplitude of $20cm$ with same period along the same line about the same equilibrium position. The maximum distance between the two is $20cm$ . Their phase difference in radians is :

• A. $\frac{2\pi }{3}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is C $\frac{\pi }{3}$

$\begin{array}{c}x=A\sin \left(\omega t\right)and\\ x=A\sin \left(\omega t+\phi \right)\end{array}$

Where $A=$ amplitude$=20$ cm for both the particles

$\omega$=angular velocity=Same for both the particles (because period is same)

$\phi$=phase difference.

Distance between the two particles is given by;

$\begin{array}{c}d=A\sin \left(\omega t+\phi \right)-A\sin \left(\omega t\right)\\ =d=A\left[2\sin \frac{\phi }{2}\right]\cos \left(\frac{2\omega t+\phi }{2}\right)\end{array}$

Clearly we will get a maximum value when $\cos \left(\frac{2\omega t+\phi }{2}\right)$=maxiumum=1

Hence,

$\begin{array}{c}d=2A\left[\sin \frac{\phi }{2}\right]\\ =20=2\times 20\left[\sin \frac{\phi }{2}\right]\\ =\frac{\phi }{2}=\frac{\pi }{6}\\ \phi =\frac{\pi }{3}\end{array}$

Hence the phase difference between the two particles will be $\frac{\pi }{3}$/