Simple Harmonic Motion Questions

[vinnie4]

Simple Harmonic Motion Questions-springs

Homework Statement

A .20 kg mass is sliding on a horizontal frictionless air track with a speed of 3 m/s when it instantaneously hits and sticks to a 1.3 kg mass initially at rest on the track. The 1.3 kg mass is connected to one end of a massless spring which has a spring constant of 100 Newtons per meter. The other end of the spring is fixed.
After the collision the two masses undergo simple harmonic motion about their position at impact.

a. Determine the amplitude of the harmonic motion.

b. Determine the period of the harmonic motion

2.An ideal spring of unstretched length .20m is placed horizontally on a frictionless table. One end of the spring is fixed and the other end is attached to a block of mass M = 8.0kg. The 8.0kg block is also attached to a massless string that passes over a small frictionless pulley. A block of mass m = 4.0kg hangs from the other end of the spring. When this spring-and-blocks system is in equilibrium, the length of the spring is .25m and the 4.0kg block is .70m above the floor.

a. Calculate the force constant k of the spring
b. Calculate the potential energy of the spring.

The string is now cut at point P ( a point along the frictionless surface beween the 8 kg mass and the pulley.)
c. What is the amplitude of the SHM of the 8 kg block?
d. Calculate the max acceleration experienced by the 8 kg block.
e. Calculate the frequency of oscillation of the 8 kg block.
f. Calculate the max speed attained by the 8 kg block

Homework Equations

KE= 1/2mv^2
UPE=1/2kx^2
T=2pi$$\sqrt{}m/k$$
Fspring = -kx
f= 1/T

The Attempt at a Solution

For 1:
a. I already had the max velocity as .4 and since the total mechanical energy equals both 1/2mv_max^2 and 1/2kA^2 I set them equal to each other and got A=.048 m

b. I used T=2pi$$\sqrt{}m/k$$ which gave me T=.769

2.
a. F=-kx
No acceleration, $$\Sigma$$F=0
F_tension=mg=40
40=-k(.05)
k=800

b.
Us=1/2kx^2
Us=1 J

c. Would the amplitude just be .05m? because the spring's unstretched length is .20 and it was stretched by .05

d.$$\Sigma$$F=Ma
Only force on block is spring force which equaled 40
40=8a
a=5m/s^2

e.T=2pi$$\sqrt{}m/k$$
T=.628

f= 1/T
f=1.59

f. 1/2kA^2=1/2mv_max^2
.5 m/s =v_max

 

[anathema]

Hello! Great job on your solutions for the first problem. Here are my thoughts on your solutions for the second problem:

a. Your calculation for the force constant looks good.

b. For the potential energy, it should be U = 1/2kx^2, where x is the displacement from the equilibrium position (not the unstretched length of the spring). So the potential energy at equilibrium would be U = 1/2k(0)^2 = 0 J.

c. The amplitude would indeed be 0.05m, as it is the displacement from the equilibrium position.

d. Your calculation for the maximum acceleration looks good.

e. Your calculation for the period looks good.

f. To find the maximum speed, we can use the energy conservation equation: KE + PE = constant. At the equilibrium position, all of the energy is potential energy, so at the maximum speed, all of the energy is kinetic energy. Therefore, 1/2mv_max^2 = 1/2kA^2. Solving for v_max, we get v_max = A*sqrt(k/m) = 0.05*sqrt(800/8) = 0.5 m/s.

Great job overall! Keep up the good work.