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 193m/s and a frequency of 240Hz . The amplitude of the standing wave at an antinode is 0.440cm .

Part A

Calculate the amplitude at point on the string a distance of 25.0cm 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 = 193/240 = 72.2 cm.

The amplitude of vibration y follows a sine curve:
y = Asin(2?.x/L)
where A=0.440 cm is maximum amplitude (ie at an antinode), x=25 cm is distance along string, and L=72.2.5 cm is wavelength.

y=0.95cm

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 =1/240Hz = 0.004 s = 4 ms.
So the time taken = 2 ms

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