Question

A particle performing linear shm with period 6 second is at positive extreme position at t= 0 . The particle is found to be at distance of 3cm from this position at t= 7 second before reaching to mean position find its amplitude? solutions

Answer 1

Answer:

Let the two points be A and B.

The particle first passes through A and then through B.

Phase at point A is

6

5π

and at B is

6

7π

.

x=Asinωt

⇒ωt

1

=

6

5π

andωt

2

=

6

7π

⇒t

2

−t

1

=

6ω

2π

=

6

1

(

ω

2π

)=

6

T

=1 s

Answer 2

Answer:

x=A sin (omegat + π/2)

Since particle starts (t=0) from positive extreme position, phi= π/2 and x=A-3

therefore, Asin (omegat+π/2)

A-3= Asin (2π/T+π/2)

A-3/A= sin (2π/6×7 +π/2)

A-3/A= sin (7π/3+π/2)

A-3/A =sin (π/3+π/2)

= cos π/3 = 1/2

therefore, 2A-6 =A

therefore, A= 6cm

therefore Amplitude of S.H.M=6cm