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# Harmonic Frequencies and spring problems

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1. If the second harmonic of a guitar string is at a frequency of 640 Hz, what is the frequency of the third harmonic?

2. A clarinet is an instrument which acts like an "open-closed" tube. Suppose you play a note which is the fundamental of the instrument (with all holes closed) which is at 415 Hz. What is the frequency of the next harmonic you can play?

3. You shine white light through a cyan filter onto a magenta sheet of paper. What color does the paper appear to be?

4. Choose the appropriate resultant color. If no light is the result, choose none.
a. Yellow light shines on a magenta object. What color is the reflected light?
b. Yellow light shines on a blue surface. What color is the reflected light?
c. Cyan light shines on a blue filter. What color is the transmitted beam?

5. a. In order to stretch a spring 0.1m from equilibrium, you have to apply a force of 2N. What is the spring constant of the spring?
b. Suppose the spring is stretched such that the displacement from equilibrium is doubled. What force is required to do this?

6. A professor drives off with his car (mass 840 kg), but forgot to take his coffee mug (mass 0.41 kg) off the roof. The coefficient of static friction between the mug and the roof is 1.3, and the coefficient of kinetic friction is 0.4. What is the maximum acceleration of the car, so the mug does not slide off?

https://brainmass.com/physics/equilibrium/harmonic-frequencies-spring-problems-520452

## SOLUTION This solution is FREE courtesy of BrainMass!

1. Given that fn = n*f1
the values for n = 1,2,3,4,5 ...
so, f2 = 2*f1 = 640 Hz
then f1 = 640 Hz / 2 = 320 Hz
Solving for f3 = 3*f1 = 3*(320 Hz) = 960 Hz

2. For clarinet, n = odd harmonics , n = 1, 3,5,7 . . .
so, fn = n*f1
since the given is the fundamental which means the first value for n = 1,
f = 3*415 Hz = 1245 Hz

3. Blue

4. a. red
b. none
c. blue

5. a. Let Force = F, spring constant = k, displacement = x
then, using F = kx
k = F/x = 2N / 0.1 m = 20 k/s/s
b. F= kx
then F = k (2x) = (20 k/s/s )*(2*0.1 m)= 4 N

6. Let u = kinetic friction
F = m*a = m*g*u
a = u*g = 1.3*(9.8 m/s/s) = 12.74 m/s/s

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