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.
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?
1 answer
1 answer