Physics, asked by rohanshriramjwar2000, 1 year ago

if two s.h.m are represented by equation y1=10sin[3πt+π÷4] and y2 =5 [sin(3πt)+√3cos(3πt)] then find ratio of their amplitudes and phase difference in between them​

Answers

Answered by sonuvuce
13

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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Answered by 0000biswajitaich
0

Answer:

1:1

EXPLANATION

In Y1 A= 10

Y2 =5 sin3πt + 5√3 cos3πt

=5 sin3πt + 5√3 sin(π/2 -3πt)

=5 sin3πt - 5√3 sin(3πt - π/2)

A net= √(5^2 + (5√3)^2 + 2×5×5√3cosπ/2)

= √(100 + 0)

= 10

so, ratio of A of Y1/Y2 =10/10

=1:1

...THANK YOU...

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