Question

A particle is subjected to two simple harmonic motions, one along the X-axis and the other on a line making an angle of 45° with the X-axis. The two motions are given by x = x0 sin ωt and s = s0 sin ωt. Find the amplitude of the resultant motion.

Answer 1

The amplitude of the resultant motion is:

• The two motions are given by ;

x = x0 sin ωt and s = s0 sin ωt

• Angle between both motion = 45°

• Resultant, R = √( x^2 + s^2 +

2.x.s.sin45°)

= √[ (x0.sinωt )^2 + (s0.sinωt )^2 +

2.x0sinωt.s0sinωt.1/√2 ]

= √( x0^2.sin^2ωt + s0^2sin^2ωt +

√2.x0.s0.sin^2ωt )

= {√(x0^2 + s0^2 +√2.x0.s0)}×sinωt

• Therefore, amplitude of resultant

motion is √(x0^2 + s0^2 +√2.x0.s0)

Answer 2

Explanation:

• Let us consider that the other line making the angle 45° with the x axis to be y axis.
• Therefore, along x axis, the particle has the equation of motion $x=x_{0} \sin \omega t$  and along y axis, the equation of motion is $s=s_{0} \sin \omega t$.  
• These two simple harmonic motions are separated by the angle $\theta=45^{\circ}$.  
• The total displacement equation can be written as $R=\sqrt{x^{2}+s^{2}+2 x s \cos 45^{\circ}}=\left[x_{0}^{2}+s_{0}^{2}+\sqrt{\left(2 x_{0} s_{0}\right)}\right]^{\frac{1}{2}} \sin (\omega t)$.  

• Hence the amplitude of the resultant motion can be written as $R=\left[x_{0}^{2}+s_{0}^{2}+\sqrt{\left(2 x_{0} s_{0}\right)}\right]^{\frac{1}{2}}$ since  $\sin (\omega t)=1$.