Question

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 193 m/s and a frequency of 215 Hz . The amplitude of the standing wave at an antinode is 0.390 cm .

Part A

Calculate the amplitude at point on the string a distance of 17.0 cm from the left-hand end of the string. (m)

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? (s)

Part C

Calculate the maximum transverse velocity of the string at this point. (m/s)

Part D

Calculate the maximum transverse acceleration of the string at this poin (m/s^2)

pls halp.

Answer 1

First calculate the wavelength using
speed = frequency x wavelength
So wavelength = speed/frequency = 193/215 = 0.8976m = 89.76 cm.

The amplitude of vibration y follows a sine curve:
y = Asin(2π.x/L)
where A=39 cm(I guess you have mentioned wrong unit in the question) is maximum amplitude (ie at an antinode), x=17 cm is distance along string, and L=89.76 cm is wavelength.
y = 39sin(2π.17/89.76) =39*0.9279 = 36.1881cm

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 =1/215Hz = 0.00465s = 4.65 ms.
So the time taken = 4.65ms.

Calculations maybe faulty, logic is correct.

Hope this helps :)

Not sure about the last two parts.