Question

A particle is subjected to two simple harmonic motions of same time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if the phase difference between the motions is (a) 0°, (b) 60°, (c) 90°.

Answer 1

It is given that a particle is subjected to two S.H.M.s of same time period in the same direction.

Amplitude of first motion, A₁ = 3 cm
Amplitude of second motion, A₂ = 4 cm

Let ϕ be the phase difference.

The resultant amplitude $\left(R\right)$ is given by,
$R=\sqrt{A_1^2+A_2^2+2A_1{A}_{2}cos\varphi }$

(a) When ϕ = 0°
$$\begin{gathered} R=\sqrt{\left(3^2+4^2+\left(2\right)\left(3\right)\left(4\right)\cos 0°\right)} \\ =7\mathrm{cm} \end{gathered}$$

(b) When ϕ = 60°
$$\begin{gathered} R=\sqrt{3^2+4^2+\left(2\right)\left(3\right)\left(4\right)\cos 60°} \\ =\sqrt{37}=6.1\mathrm{cm} \end{gathered}$$

(c) When ϕ = 90°
$$\begin{gathered} R=\sqrt{\left(3^2+4^2+\left(2\right)\left(3\right)\left(4\right)\cos 90°\right)} \\ =\sqrt{25}=5\mathrm{cm} \end{gathered}$$