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 195 m/s and a frequency of 200 Hz . The amplitude of the standing wave at an antinode is 0.350 cm .A. Calculate the amplitude at point on the string a distance of 25.0 cm from the left-hand end of the string? B How much time does it take the string to go from its largest upward displacement to its largest downward displacement at this point? C Calculate the maximum transverse velocity of the string at this point? D Calculate the maximum transverse acceleration of the string at this point?

Explanation / Answer

What we mostly need is the length of the string.
We can get this from the equation
f(n) = nv/2L,
where f(n) is the frequency of the nth harmonic.
In the case at hand, n=3, f(n)=200 Hz, and v = 195 m/s;
therefore, L = 3 (195 m/s) / (2 * 200 s^(-1)) = 1.46 m = 146 cm

So there are three arcs of 48.66 cm length and 0.35 cm maximum deflection.
If y = 0.35 cm sin(pi x / 48.66 cm) when the wave is as "bent" as possible,
then at x = 25.0 cm, we have
A) y = 0.35 cm sin(25 pi/48.66) = 0.349 cm at that time.
The 0.349 cm is the amplitude at the point x = 25 cm

B) 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

At the antinode y=Asinwt, Transverse speed = dy/dt=Awcoswt = 0.349 *2 *3.14 *200 cos(2 *3.14 *200 * 2.5 x10^-3) = 437.68 m/s

Transverse acceleration = dy^2/dt^2=-Aw^2(sinwt) =
The maximum acceleration is when sinwt =1 therefore, maximum acceleration=Aw^2 = 0.349 *(2*3.14*200)^2=550560.064m/s2

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