Question

⭕ 50 points ⭕

Two SHMs S1=(a sin omega t )and S2 = (bsin omega t) are superimposed...

Explain ....!!!​

Answer 1

Given:

Two SHMs S1=(a sin omega t )and S2 = (bsin omega t) are superimposed

To Find:

1. If the particle performs SHM.
2. It travels in straight line.

Solution:

Given S2 is at an angle of 37° to S1.

Therefore lets assume X be the total displacement in direction of S1 and Y be perpendicular to S1.

Then,

• X = S1 + S2 cos 37° = asin ωt + 0.8b sin ωt -- (a)
• Y = S2 sin 37° = 0.6b sin ωt -- (b)

Motion of the particle will be along : Xi + Yj

Net Motion:

• $\sqrt{a^{2} + b^{2} }$ sin ωt , is of the form A sin ωt
• Hence the motion is Simple Harmonic.

Now ,

from (b) ,

• sin ωt = Y/0.6b

Substitute value of sin ωt in  ( a ) ,

• X = (a + 0.8b ) Y / 0.6b
• Therefore,
• X =  k Y , k = (a + 0.8b)/0.6b
• Motion will be along straight line.

Hence the particle will perform SHM as well as travel in straight line.

Answer 2

Answer:

Two SHMs S1=(a sin omega t )and S2 = (bsin omega t) are superimposed

To Find:

If the particle performs SHM.

It travels in straight line.

Solution:

Given S2 is at an angle of 37° to S1.

Therefore lets assume X be the total displacement in direction of S1 and Y be perpendicular to S1.

Then,

X = S1 + S2 cos 37° = asin ωt + 0.8b sin ωt -- (a)

Y = S2 sin 37° = 0.6b sin ωt -- (b)

Motion of the particle will be along : Xi + Yj

Net Motion:

\sqrt{a^{2} + b^{2} }a2+b2 sin ωt , is of the form A sin ωt

Hence the motion is Simple Harmonic.

Now ,

from (b) ,

sin ωt = Y/0.6b

Substitute value of sin ωt in  ( a ) ,

X = (a + 0.8b ) Y / 0.6b

Therefore,

X =  k Y , k = (a + 0.8b)/0.6b

Motion will be along straight line.

Hence the particle will perform SHM as well as travel in straight line.