Physics - SHM Question with Solution | TestHub

Assuming the two pendulums follow the following functions of their displacement :

θ₁= θ₀sin (ωt + ϕ₁)             ..(i)

and                      θ₂= θ₀sin (ωt + ϕ₂)             ..(ii)

As it is given that amplitude and time period being equal but phases being different.

Now, for first pendulum at any time t

θ₁= θ₀(Right extreme)

From Eq. (i), we get

⇒                      θ₀= θ₀sin (ωt + ϕ₁)

or                      1 = sin (ωt + ϕ₁)

or$\left(\omega t+{\varphi }_1\right)=\frac{\pi }{2}\dots \left(\text{iii}\right)$

Similarly, at the same instant t for pendulum second, we have

${\theta }_2=-\frac{{\theta }_0}{2}$

Noteθ₀= 2^ois the angular amplitude for the second pendulum the angular displacement is one degree therefore,${\theta }_2=\frac{{\theta }_0}{2}$and negative sign is taken to show for being left to mean position

From Eq. (ii), then

$-\frac{{\theta }_0}{2}={\theta }_0\text{sin}\left(\omega t+{\varphi }_2\right)$

$\Rightarrow \text{sin}\left(\omega t+{\varphi }_2\right)=-\frac{1}{2}$

$\Rightarrow \left(\omega t+{\varphi }_2\right)=-\frac{\pi }{6}\text{or}\frac{7\pi }{6}$

$\text{or}\left(\omega t+{\varphi }_2\right)=-\frac{\pi }{6}\text{or}\frac{7\pi }{6}\dots \left(\text{iv}\right)$

From Eqs. (iv) and (iii), the difference in phases

$\left(\omega t+{\varphi }_2\right)-\left(\omega t+{\varphi }_1\right)=-\frac{7\pi }{6}\text{or}\frac{\pi }{2}=\frac{7\pi -3\pi }{6}=\frac{4\pi }{6}$

$\text{or}\left({\varphi }_2-{\varphi }_1\right)=\frac{4\pi }{6}=\frac{2\pi }{3}=120^o$