A 0 050 kg object is attached to a horizontal spring and placed on a frictionles

Question

A 0 050 kg object is attached to a horizontal spring and placed on a frictionless surface. It is stretched 10.0 cm and then released. It will execute a simple harmonic motion. It was found that its period Tis 0 60 sec. Its displacement x from the equilibrium location is given by x = A cos t where A is the amplitude of the motion in m, is time measured in seconds and o is angular velocity in rad/s a. What is the value of the amplitude A? b. Compute the spring constant k of the spring c. Compute the frequency f d. Compute the value of o. e. Plot a graph of displacement x and elastic potential energy as a function of time t from 0 sec to 0.6 sec. f When is the displacement x greatest?

Explanation / Answer

Mass m = 0.05 kg , strech in the spring is x = 0.1 m ,  

Time period is T = 0.6 s

x = A cos W*t

we know that the time period is T = 2pi sqrt(m/k)

T^2 = 4 pi^2 (m/k)

k = 4 pi^2 (m/T^2)

substituting the values we get k = 4 pi^2(0.05/0.6^2) N/m= 5.483 N/m

and W = 2pi/T = 2pi/0.6 rad/s = 10.47 rad/s

the amplitude is the maximum stretch A = 0.1 m

b. k = 5.483 N/m

c. frequency f, W = 2pi*f ==> f = W/2pi = 10.47/2pi Hz = 1.67 Hz

d. W = 10.47 rad/s

f .

x = A cos wt

at t = 0 , x= A = 0.1 m

at t= 0.1 s , X = 0.1*cos(10.47*0.1) m= 0.0999833 m

at t= 0.1 s , X = 0.1*cos(10.47*0.2) m= 0.09993322m

at t= 0.1 s , X = 0.1*cos(10.47*0.3) m= 0.0998497716686 m

at t= 0.1 s , X = 0.1*cos(10.47*0.4) m= 0.0997329794323 m

at t= 0.1 s , X = 0.1*cos(10.47*0.5) m= 0.0995828848516 m

at t= 0.1 s , X = 0.1*cos(10.47*0.6) m= 0.0993995380455m

f. the displacement will be maximum at extream end at t= 0 s , amplitude = displacemenr = 0.0999833 m

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