Question

A toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/m . When the toy is 0.140 m from its equilibrium position, it is observed to have a speed of 3 m/s and a total energy of 5.2 J .

a) find the mass of the toy in kg

b) find the amplitude of the motion

c) find the max speed attained by object during its motion

Answer 1

Part A

Total energy in SHM at any point is given by:

TE = KE + PE

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

k = 300 N/m

x = 0.140 m

TE = 5.2 J

V = 3 m/sec

5.2 = 0.5*m*3^2 + 0.5*300*0.140^2

m = (5.2 - 150*0.140^2)/4.5

m = 0.502 kg

Part B

Since total energy is constant, So at the point where velocity is zero, total energy will be

TE = PEmax = 0.5*k*A^2

PEmax = 5.2 J

A = sqrt (2*PEmax/k)

A = sqrt (2*5.2/300)

A = Amplitude = 0.186 m

Part C

angular velocity will be given by:

w = sqrt (k/m)

w = sqrt (300/0.502)

w = 24.45 rad/sec

Max Velocity is given by:

Vmax = A*w

Vmax = 0.186*24.45 = 4.55 m/sec

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