Question

A toy of mass 0.150 kg is undergoing simple harmonic motion (SHM) on the end of a horizontal spring with force constant 300 N/m . When the object is a distance 1.25×10?2 m from its equilibrium position, it is observed to have a speed of 0.305 m/s .

a) What is the total energy of the object at any point of its motion?

b) What is the amplitude of the motion?

c) What is the maximum speed attained by the object during its motion?

Answer 1

Total energy at any point will be given by:

TE = KE + PE

TE = 0.5*m*V^2 + 0.5*k*x^2

at given point x = 1.25*10^-2 m, TE will be

TE = 0.5*0.150*0.305^2 + 0.5*300*(1.25*10^-2)^2

TE = 0.0304 J

Part B

from energy conservation we know that at max distance from equilibrium position (Amplitude), TE will be equal to max Potential Energy

TE = PE_max

TE = 0.5*k*A^2

A = sqrt (2*TE/k)

A = Amplitude

A = sqrt (2*0.0304/300) = 1.42*10^-2 m

Part C

Max speed is given by:

Vmax = A*w

w = sqrt (k/m)

Vmax = A*sqrt (k/m)

Vmax = 1.42*10^-2*sqrt (300/0.15)

Vmax = 0.635 m/sec

Please Upvote.