Question

If two S.H.M. are represented by equations y¹ = 10 sin 3t+
and y²=5 sin(31) + V3 cos(3rt) then find the ratio of their amplitudes and phase difference in between them.​

Answer 1

The ratio of the amplitudes is 1:1

The phase difference is π/12

Explanation:

We know that the general equation of SHM

$\boxed{y=A\sin(\omega t+\phi)}$

Where, A is the amplitude, $\omega$ is angular frequency and $\phi$ is phase difference

Given

The first equation

$y_1=10\sin(3\pi t+\frac{\pi}{4}$

Comparing this with the general equation

We get

Amplitude $A_1=10$ units

Phase difference $\phi_1=\frac{\pi}{4}$

The second equation

$y_2=5[\sin(3\pi t)+\sqrt{3}\cos(3\pi t)]$

or, $y_2=10[\frac{1}{2}\sin(3\pi t)+\frac{\sqrt{3}}{2}\cos(3\pi t)]$

or, $y_2=10[\cos(\frac{\pi}{3})\sin(3\pi t)+\sin(\frac{\pi}{3})\cos(3\pi t)]$

or, $y_2=10\sin(3\pi t+\frac{\pi}{3})$

Comparing the above with the general equation of SHM

Amplitude $A_1=10$ units

Phase difference $\phi_2=\frac{\pi}{3}$

Ratio of the amplitudes

$\frac{A_1}{A_2}=\frac{10}{10}=1$

or, $A_1:A_2=1:1$

Phase difference between the two

$=\phi_2-\phi_1$

$=\frac{\pi}{3}-\frac{\pi}{4}$

$=\frac{\pi(4-3)}{12}$

$=\frac{\pi}{12}$

Hope this answer is helpful.

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