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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A 0.5 kg mass is hung on a vertical massless spring. The new equilibrium position of the spring is found to be 3 cm below the equilibrium position of the spring without the mass. a) What is the spring constant, k ? b) Show that the mass and spring system oscillates with simple harmonic motion about the new equilibrium position.


Since the mass/spring system is in equilibrium, the downward force of gravity must be balanced by the upward pull of the spring. See diagram below (Note that we have for convenience defined the positive x-axis to point downwards.).
Figure 13.5: Problem 13.1
\epsfxsize=2 in

Thus at x = x  eq = 0.03m ,

$\displaystyle\sum$F = 0 = kx  eq - mg (18)

$\displaystyle\Rightarrow$ k = $\displaystyle{\frac{mg}{x_{\;{\:\rm eq}}}}$   
  = $\displaystyle{\frac{(0.5)(9.8)}{(0.03)}}$ = 163  N/m .   
If the spring is now displaced a distance s above the new equilibrium position, so that x = x  eq - s , the net force upward due to the spring is:

Fs = k(x  eq - s)

The total force upward is therefore

F  tot = Fs - mg = kx  eq - ks - mg

Since the equilibrium position obeys Eq.(13.18), the equation for the acceleration of the mass reduces to:

F  tot = - ks (19)

i.e. The net force obeys Hook's law as a function of the displacement, s , and the mass undergoes simple harmonic motion about the new equilibrium position. Note that Eq.(13.19) is valid for both positive and negative values of s .

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Consider a SHO with m = 0.5 kg, k = 10 N/m and amplitude A = 3 cm. a) What is the total energy of the oscillator? b) What is its maximum speed? c) What is the speed when x = 2 cm? d) What are the kinetic and potential energies when x = 2 cm?



E total = $\displaystyle{\textstyle\frac{1}{2}}$kA 2 = $\displaystyle{\textstyle\frac{1}{2}}$(10)(0.03) = 0.0045  J


v max = $\displaystyle\sqrt{\frac{kA^2}{m}}$ = $\displaystyle\sqrt{\frac{(10)(0.03)^2}{(0.5)}}$ = 0.134  m/s

Using Eq.(13.7) for the velocity at x = 0.02m ,
v = $\displaystyle\sqrt{\frac{k(A^2-x^2)}{m}}$   
  = $\displaystyle\sqrt{\frac{(10)((0.03)^2 - (0.02)^2)}{(0.5)}}$ = 0.10  m/s   
The potential and kinetic energies at that point are:
PEs = $\displaystyle{\textstyle\frac{1}{2}}$kx 2 = $\displaystyle{\textstyle\frac{1}{2}}$(10)((0.02)2) = 0.002  J   
KE = $\displaystyle{\textstyle\frac{1}{2}}$mv 2 = $\displaystyle{\textstyle\frac{1}{2}}$(0.5)((0.1)2) = 0.0025  J   

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

If the oscillator in the above problem is released from rest at x = A when the clock is set to t = 0 seconds, a) determine the position and velocity of the oscillator at t = 2 s. b) At what time does the oscillator get to x = - 1 cm?


At t = 2 s, we have:
x(t) = Acos $\displaystyle\left( \sqrt{\frac{k}{m}}t \right)$   
  = (0.03)cos $\displaystyle\left( \sqrt{\frac{10}{0.5}} (2) \right)=$ - 0.027  m   
v(t) = $\displaystyle\sqrt{\frac{k}{m}}$Asin $\displaystyle\left( \sqrt{\frac{k}{m}}t \right)$   
v(2) = $\displaystyle\sqrt{\frac{10}{0.5}}$(0.03)sin $\displaystyle\left( \sqrt{\frac{10}{0.5}} (2) \right)=$0.062  m/s   
We must find the value of t for which the displacement
x(t) = Acos $\displaystyle\left( \sqrt{\frac{k}{m}}t = -1\;{\:\rm cm} \right)$   
$\displaystyle\Rightarrow$ t = $\displaystyle\sqrt{\frac{m}{k}}$$\displaystyle\cos^{-1}_{}$$\displaystyle\left( \frac{x}{A}
  = $\displaystyle\sqrt{\frac{0.5}{10}}$$\displaystyle\cos^{-1}_{}$$\displaystyle\left( \frac{-0.01}{0.03}
\right)=$0.43  s   

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

If the mass in the above problem experiences a constant frictional force of 0.1 N and is released from rest at x = A , a) what is the velocity of the mass when it first passes through the equilibrium position at x = 0 . b) At what value of x does the velocity of the mass first go to zero after it has been released?


When the mass passes through the equilibrium position, it has covered a distance of d = 0.03 m. Applying the Work-Energy Theorem to the problem:
W ext = $\displaystyle\Delta$KE + $\displaystyle\Delta$PEs + $\displaystyle\Delta$PE grav   
- F frd = $\displaystyle\left( {1\over 2} m v_{f}^{2} - 0 \right)+$$\displaystyle\left( 0
- {1\over 2} k
A^2 \right)+$ 0   
$\displaystyle\Rightarrow$ vf = $\displaystyle\sqrt{\frac{-2 F_{\:\rm fr} d + kA^2}{m}}$   
  = $\displaystyle\sqrt{\frac{-2 (0.1)(0.03) + (10)(0.03)^2}{0.5}}$ = 0.077  m/s   
Assume that the velocity of the mass vanishes at x = x0 . The total distance moved from its starting position is therefore d = A - x0 and the Work-Energy Theorem states
W ext = $\displaystyle\Delta$KE + $\displaystyle\Delta$PEs + $\displaystyle\Delta$PE grav   
- F frd = 0 + $\displaystyle\left( {1\over 2} k x_0^{2} - {1\over 2} k A^2 \right)+$ 0   
- F fr(A - x0) = $\displaystyle\left( {1\over 2} k x_0^{2} - {1\over 2} k A^2 \right)$   
$\displaystyle\Rightarrow$ 0 = kx02 - 2F fricx0 - kA 2 + 2F fric   
  = (10)x02 - 2(0.1)x0 + $\displaystyle\left[ -(10)(0.03^2) + 2(0.1)(0.03)
  = (10)x02 - 0.2x0 - 0.003   
This is a quadratic equation which we can solve for the unkown, x0 . The two possible answers are
x0 = $\displaystyle{0.2\pm\sqrt{(0.2)^2-4\cdot 10 \cdot (-0.003)}\over
2\cdot 10}$  m   
  = - 0.01  m   or 0.03  m (20)
The solution x0 = 0.03 m corresponds to the original starting position, at which the mass did indeed have zero velocity. It is not the answer we are interested in, however. The physically relevant answer is x0 = - 0.01 Thus, the mass comes to rest at 0.01 m past the original equilibrium position.

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a stopwatch, the student determines that the period of the pendulum is 1.44 s. Determine the experimental value of the gravitational acceleration.

Since the angle through which the pendulum is initially displaced is small, Eq.(13.17) in the Lecture Notes is valid for the period T :

T = 2$\displaystyle\pi$$\displaystyle\sqrt{\frac{L}{g}}$   
$\displaystyle\Rightarrow$ g = $\displaystyle{\frac{4 \pi^{2} L}{T^2}}$   
  = $\displaystyle{\frac{4 \pi^{2} (0.510)}{1.44^2}}$ = 9.71 m/s 2   
which is within 1% of the correct value 9.8 m/s 2.

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