A uniform string fixed at one end passes over a pulley and then is attached to a hanging object. The string is horizontal between the fixed end and the pulley. When you let the hanging object oscillate in the vertical plane, the fundamental frequency of the horizontal part of the string also oscillates. If the sound of the string changes by one octave during the oscillation, what maximum angle does the oscillating part of the string make to the vertical? (Hint: When a frequency is doubled, the sound changes by an octave.)

Would I have to calculate the force of the pulley and the object in order to do this problem? How would I do it?

For the fundamental frequency to change by an octave, the tension T in the string must vary by a factor of 4. (Wave speed and fundamental frequency vary with the square root of the tension).

Consider how large an amplitude the pendulum must swing to have the tension vary by a factor of four. The largest tension will occur at the bottom of the swing, when the speed and centripetal force are highest. The least tension will occur at the maximum swing angle, A.

Maximum tension:
Tmax = M g + M Vmax^2/L
(1/2)Vmax^2 = g L(1 - cos A)
Tmax = M g [1 + 2 (1 - cosA)]
= M g [3 - 2 cos A]
Minimum tension occurs at maximum angle A, when there is no centripetal acceleration, and is
Tmin = M g cos A

Setting Tmax/Tmin = 4 will let you finish the problem and solve for the angle A.

Check my thinking. I may have made a math error along the way.

To solve this problem, you do not need to calculate the force of the pulley or the object. The key to solving this problem lies in understanding the relationship between the frequency and the length of a string.

When a string vibrates, its fundamental frequency is determined by its length, tension, and mass per unit length. The fundamental frequency is the lowest frequency at which the string can vibrate.

In this case, the oscillation of the hanging object causes the horizontal part of the string to also vibrate. When the sound of the string changes by one octave, it means the frequency doubles. Since the length and tension of the string remain constant, the only variable that can change is the mass per unit length of the string.

Now, when a string vibrates in a standing wave pattern, its harmonics or overtones are integer multiples of the fundamental frequency. The second harmonic has twice the frequency of the fundamental, the third harmonic has three times the frequency, and so on. When the frequency of the fundamental doubles, it means that the entire series of harmonics double as well.

This tells us that the oscillating part of the string, between the fixed end and the pulley, must form the second harmonic. In a string fixed at one end and vibrating in the standing wave pattern, the second harmonic has one antinode and two nodes.

The antinode is the point of maximum displacement, and it always occurs halfway between the two nodes. In this case, the antinode represents the maximum displacement of the oscillating part of the string.

Since the length of the oscillating part of the string is equal to the distance between the fixed end and the pulley, the maximum angle that the oscillating part of the string makes with the vertical is determined by the length of this section. Specifically, it is the angle between the two segments of the horizontal string on either side of the pulley.

To find this angle, you can use the geometry of the situation. Suppose the distance between the fixed end and the pulley is "L" units. Then, the oscillating part of the string is also "L" units long.

Since the antinode occurs at the halfway point of this length (i.e., L/2), the maximum angle can be determined using trigonometry:

sin(theta) = opposite / hypotenuse = (L/2) / L = 1/2

Taking the inverse sine (sin^(-1)) of both sides gives:

theta = sin^(-1)(1/2)

Calculating the value of sin^(-1)(1/2), we find:

theta = 30 degrees

Therefore, the maximum angle that the oscillating part of the string makes with the vertical is 30 degrees.