A 100 g mass is undergoing simple harmonic motion along the x-axis on an ideal spring with k=150 N/m. At t = 0, the mass is at x = -10 cm and is moving with velocity v = 15 m/sec. a) What is the total energy of the system? b) What is the position of the mass as a function of time, x(t)? c) What is the acceleration of the mass at t = 2.5 sec?

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A 100 g mass is undergoing simple harmonic motion along the x-axis on an ideal spring with k=150 N/m.
At t = 0, the mass is at x = -10 cm and is moving with velocity v = 15 m/sec.
a) What is the total energy of the system?
b) What is the position of the mass as a function of time, x(t)?
c) What is the acceleration of the mass at t = 2.5 sec?
A system with the same mass and spring constant as the question above has a damping constant
y = 1.1. Assume the mass was pulled to the right 30 cm at t = 0 and released.
a) Estimate the time at which the amplitude has decayed to 4 of its initial value.
b) Assume the system is connected to a forcing function given by (in Newtons) F(t) = 3 cos wt Estimate
the value of the amplitude at resonance.
Transcribed Image Text:A 100 g mass is undergoing simple harmonic motion along the x-axis on an ideal spring with k=150 N/m. At t = 0, the mass is at x = -10 cm and is moving with velocity v = 15 m/sec. a) What is the total energy of the system? b) What is the position of the mass as a function of time, x(t)? c) What is the acceleration of the mass at t = 2.5 sec? A system with the same mass and spring constant as the question above has a damping constant y = 1.1. Assume the mass was pulled to the right 30 cm at t = 0 and released. a) Estimate the time at which the amplitude has decayed to 4 of its initial value. b) Assume the system is connected to a forcing function given by (in Newtons) F(t) = 3 cos wt Estimate the value of the amplitude at resonance.
Expert Solution
Step 1

(a)

Given:

The mass attached to the spring is 100 g.

The force constant of the ideal spring is 150 Nm.

The position of mass is -10 cm.

The velocity of the mass is 15 ms.

Introduction:

Simple harmonic motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position.

Step 2

Calculation:

Write the expression of the total energy of the system.

E=12mv2+12kx2

Substitute 100 g for m15 msec for v150 Nm for k and -0.1 m for x in the above expression.

E=120.1 kg15 msec2+12150 Nm-0.1 m2E=11.25 J+0.75 JE=12 J

Thus, the total energy of the system is 12 J.

(b)

Calculation:

Write the expression of angular velocity.

ω=km

Substitute 150 Nm for k and 0.1 kg for m in the above expression.

ω=150 Nm0.1 kgω=38.7 rads

Write the expression of xt.

xt=Acosωt

Substitute 0 for t in the above expression.

-0.1 m=Acosω0A=-0.1 m

Thus, the expression of xt=-0.1 mcos38.7t.

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