Question 12 Marks
Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of 2° to the right with the vertical, the other pendulum makes an angle of 1° to the left of the vertical. What is the phase difference between the pendulums?
Answer
View full question & answer→$\theta_1=\theta_0\sin(\omega\text{t}+\delta_1)$$\theta_2=\theta_0\sin(\omega\text{t}+\delta_2)$
For the first, $\theta=2^\circ,\therefore\sin(\omega\text{t}+\delta_1)=1$
For the 2nd, $\theta=-1^\circ,\ \therefore\sin(\omega\text{t}+\delta_2)=-1$
$\therefore\omega\text{t}+\delta_1=90^\circ,\omega\text{t}+\delta_2=-30^\circ$
$\therefore\delta_1-\delta_2=120^\circ$
For the first, $\theta=2^\circ,\therefore\sin(\omega\text{t}+\delta_1)=1$
For the 2nd, $\theta=-1^\circ,\ \therefore\sin(\omega\text{t}+\delta_2)=-1$
$\therefore\omega\text{t}+\delta_1=90^\circ,\omega\text{t}+\delta_2=-30^\circ$
$\therefore\delta_1-\delta_2=120^\circ$