A string with both ends held fixed is vibrating in its third harmonic. The waves

Question

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 192m/s and a frequency of 200Hz . The amplitude of the standing wave at an antinode is 0.390cm .

Part A: Calculate the amplitude at point on the string a distance of 19.0

cm from the left-hand end of the string.

Part B: How much time does it take the string to go from its largest upward displacement to its largest downward displacement at this point?

Part C: Calculate the maximum transverse velocity of the string at this point.

Part D: Calculate the maximum transverse acceleration of the string at this point.

Explanation / Answer

First calculate the wavelength using
speed = frequency x wavelength
So wavelength = speed/frequency = 192/200 = 0.96 m = 96 cm.

The amplitude of vibration y follows a sine curve:
y = Asin(2?.x/L)
where A=39 cm is maximum amplitude (ie at an antinode), x=19 cm is distance along string, and L=96 cm is wavelength.

y = 0.846 cm.

The time taken in moving from largest upward to largest downward displacements is half a period (the same as for any point along the string).
Period = 1/frequency =1/200Hz = 0.005 s = 5 ms.
So the time taken = 2.5 ms.

maximum speed is Aw = 0.39*10^-2 x 200 x (2pi) = 490.088
Maximum acceleration is Aw^2 = 0.39*10^-2 x (200 x 2pi)^2 = 615863.3146

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