Question

A particle is subjected to three SHMs in the same direction simultaneously each having equal amplitude ‘a’ and equal time period. The phase of the second motion is ${{30}^{0}}$ ahead of the first and the phase of the third motion is ${{30}^{0}}$ ahead of the second. Find the amplitude of the resultant motion.

Answer 1

It is given that the particle is subjected to three simple harmonic motions at the same time. The phase difference among the three motions causes the particle to undergo a complicated oscillation. We can find the total amplitude from the equation of motion of the particle.

Complete answer:
It is given that the particle undergoes a simple harmonic motion with phase differences. The first harmonic motion is supposed to be relating the other two motions. It is said that the second motion is ${{30}^{0}}$ahead of the first and third is ${{30}^{0}}$ ahead of the second. From this we can understand that the third is ${{60}^{0}}$ahead of the first motion.
We can write the solutions of motion for the three SHMs as –

& {{x}_{1}}=a\sin \omega t \\\ & {{x}_{2}}=a\sin (\omega t+{{30}^{0}}) \\\ & {{x}_{3}}=a\sin (\omega t+{{60}^{0}}) \\\ \end{aligned}$$ Now, we know that the three SHMs contribute to the particle motion. So, the displacement of the particle at any instant is the sum of the three SHMs as – $$\begin{aligned} & x={{x}_{1}}+{{x}_{2}}+{{x}_{3}} \\\ & \Rightarrow \text{ }x=a\sin \omega t+a\sin (\omega t+{{30}^{0}})+a\sin (\omega t+{{60}^{0}}) \\\ & \Rightarrow \text{ }x=a\sin \omega t+a[\cos \omega t\sin {{30}^{0}}+\cos {{30}^{0}}\sin \omega t]+a[\cos \omega t\sin {{60}^{0}}+\cos {{60}^{0}}\sin \omega t] \\\ & \Rightarrow \text{ }x=a\sin \omega t+a[\cos \omega t\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\sin \omega t]+a[\cos \omega t\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\sin \omega t] \\\ & \Rightarrow \text{ }x=a[\sin \omega t+\dfrac{\cos \omega t+\sin \omega t+\sqrt{3}\sin \omega t+\sqrt{3}\cos \omega t)}{2}] \\\ & \Rightarrow \text{ }x=a[\dfrac{\sqrt{3}+1}{2}\cos \omega t+\dfrac{\sqrt{3}+3}{2}\sin \omega t] \\\ \end{aligned}$$ The amplitude of the particle in SHM is given by – $$\begin{aligned} & A=\sqrt{\dfrac{{{a}^{2}}{{(\sqrt{3}+1)}^{2}}+{{a}^{2}}{{(\sqrt{3}+3)}^{2}}}{4}} \\\ & \Rightarrow \text{ }A=\dfrac{a}{2}\sqrt{3+2\sqrt{3}+1+3+6\sqrt{3}+9} \\\ & \Rightarrow \text{ }A=\dfrac{a}{2}\sqrt{16+8\sqrt{3}} \\\ & \Rightarrow \text{ }A=\dfrac{a}{2}2\sqrt{4+2\sqrt{3}} \\\ & \Rightarrow \text{ }A=a\sqrt{4+2\sqrt{3}} \\\ \end{aligned}$$ The amplitude of the particle subjected to three SHMs is $$A=a\sqrt{4+2\sqrt{3}}$$. **Note:** We have found the maximum amplitude of the particle by normalising the amplitude in the equation. The SHM is derived easily as the time period, frequency and amplitude of each of the individual SHMs are equal. We need to do more calculation to find out if the time period was different.