Please answer all 5 parts 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 constanty = 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 Estimatethe value of the amplitude at resonance.

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Please answer all 5 parts

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 $150 \frac{N}{m}$ .

The position of mass is -10 cm.

The velocity of the mass is $15 \frac{m}{s}$ .

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.

$E = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2}$

Substitute $100 g$ for $m$ , $15 \frac{m}{\sec}$ for $v$ , $150 \frac{N}{m}$ for $k$ and -0.1 m for $x$ in the above expression.

$E = \frac{1}{2} (0.1 kg) {(15 \frac{m}{\sec})}^{2} + \frac{1}{2} (150 \frac{N}{m}) {(- 0.1 m)}^{2} E = 11.25 J + 0.75 J E = 12 J$

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

(b)

Calculation:

Write the expression of angular velocity.

$ω = \sqrt{\frac{k}{m}}$

Substitute $150 \frac{N}{m}$ for $k$ and 0.1 kg for $m$ in the above expression.

$ω = \sqrt{\frac{150 \frac{N}{m}}{0.1 kg}} ω = 38.7 \frac{rad}{s}$

Write the expression of $x (t)$ .

$x (t) = A \cos ω t$

Substitute 0 for $t$ in the above expression.

$\begin{array}{rcl} - 0.1 m & = & A \cos (ω (0)) \\ A & = & - 0.1 m \end{array}$

Thus, the expression of $x (t) = (- 0.1 m) \cos (38.7 t)$ .

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