Single Harmonica Mocha (SHM) - Solutions

Questions:
2. Since acceleration is proportional to displacement from equilibrium, the acceleration will be zero when the displacement is zero. This happens at the equilibrium position.
7. Frequency will stay the same (look at T for a mass on a spring as an example). Doubling the amplitude will quadruple the maximum potential energy (U = .5*kx^2) which will double the maximum speed since max PE = max KE. Doubling the amplitude also double the maximum displacement, which means the acceleration is also double (a is proportional to -x). The total mechanical energy is quadrupled since there is four times as much initial potential energy.

Problems:
8. Since the cord is elastic, we can treat it like a spring and find a 'spring constant' for it. Rearranging the equation for formula of the period for a mass on a spring, k = 213.2 N/m. We can then calculate that the new frequency with the new mass is 3.77 Hz. Don't forget that frequency is 1/T!

9.
a) Velocity is max at equilibrium position.
0 + .5*k*A^2 = .5*m*v^2 + 0 at equilibrium
k for spring = 177.7 N/m from the period of a mass on a spring for this problem.
solving the first equation, v = 2.82 m/s

b) energy is conserved
0 + .5*k*A^2 = .5*m*v^2 + .5*k*x^2 with x = .10 m
v = 2.108 m/s

c)
total energy of the system is equal to the initial potential energy = .5*k*A^2 = 2 J

d) For x to be a maximum at t = 0, the function must be a cosine function.

x(t) = 0.15*cos(wt) with w = 2*pi/T and T = 1/3 of a second.

15.
The work done on the spring is equal to the potential energy stored in it:
3.0 J = 1/2*k*x^2 with x = 0.15 m

k = 416.7 N/m

The maximum acceleration occurs when the displacement is greatest. This occurs at the beginning when x = .12 m.

a = -k*x/m = -15 m/s^2, so m = 3.33 kg

31.
a)
T = 2*pi*(l/g)^(1/2) and f = 1/T = .613 Hz
b) Using conservation of energy:
0 + mg*(L - L cos(12)) = .5*m*v^2 + 0
v = .532 m/s
c) total mechanical energy is either equal to the max KE or the max PE.
Emech = .5*(.310 kg)*(.532 m/s)^2 = 0.0439 J

32. This problem is very similar to 31. Use conservation of energy, and the fact that the height above the bottom of the swing is L - L cos(theta).
mg(L - L cos(theta)) = .5*m*v^2
v = (2gL(1 - cos(theta))^(1/2)