Question

A free particle is executing simple harmonic motion in a straight line with a time period of 25 seconds; 5 seconds after it has crossed the equilibrium point, the velocity is found to be 0.7m/s. Find the displacement at the end of 10 seconds and also amplitude of oscillations.

Answer 1

Answer

ω=

T

2π

=

16

2π

=

8

π

rad/s

If at t=0, particle passes through its mean position (x=Asinωt) with maximum speed its v−t equation can be written as

v=ωAcosωt

Substituting the given values, we have

2=(

8

π

)Acos(

8

π

)(2)

∴ A=

π

16

2

m=7.2 m

Answer 2

Answer:

Displacement at the end of 10 seconds = 5.3 m

Amplitude of oscillations = A = 9.01 m

Explanation:

Given:

Time period = 25 seconds = T

Velocity at 5 seconds = 0.7 m/s

We know that , velocity = v = ωAcosωt

So, 0.7= (2π/T)*A*cos((2π/T)*5)

    0.7 = (2π/25)*A*cos((2π/25)*5)

 So, A = (0.7*25)/(2π*cos(2π/5) = 9.01 m

Now, the displacement after 10 seconds = x(10)

x(10) = Asinω(10)

       = 9.01*sin((2π/T)*10)

       = 9.01*sin((2π/25)*10)

       = 5.3 m