A 314 g object is attached to a massless spring and executes simple harmonic motion with a period of 0.200 sec. on a frictionless horizontal surface. If the total energy of the system is 4.67 Joules find the following: the maximum speed of the object a b the force constant of the spring the amplitude of the motion

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7A 314 g object is attached to a massless spring and executes
simple harmonic motion with a period of 0.200 sec. on a
frictionless horizontal surface. If the total energy of the
system is 4.67 Joules find the following:
the maximum speed of the object
a
b
the force constant of the spring
the amplitude of the motion
2
Transcribed Image Text:7A 314 g object is attached to a massless spring and executes simple harmonic motion with a period of 0.200 sec. on a frictionless horizontal surface. If the total energy of the system is 4.67 Joules find the following: the maximum speed of the object a b the force constant of the spring the amplitude of the motion 2
Expert Solution
Step 1

a)

The total energy of the system is the addition of the spring potential energy and kinetic energy.

In this case, The total energy is converted into kinetic energy.

Let m be the mass, v be the velocity, E is the total energy.

E=12mv2v2=2Em=2Em=2×4.67 J0.314 kgv=5.45 m/s

Therefore, the velocity is 5.45 m/s

 

Step 2

ω=kmSince, T=2πωT=2πkmSolve for spring constant 'k'.T2=4π2kmkm=4π2T2k =m×4π2T2Plug the known values.  =0.314 kg ×4π20.2002k=310 N/m

Therefore, the spring constant is 310 N/m

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