Standing Wave on a String: One end of a horizontal string of linear density 4.34

Question

Standing Wave on a String: One end of a horizontal string of linear density 4.34E-4 kg/m is attached to a small-amplitude mechanical 62.0 Hz vibrator. The string passes over a pulley, a distance L = 1.58 m away, and weights are hung from this end, as seen in the figure below. Assume the string at the vibrator is a node, which is nearly true.

A) What mass must be hung from this end of the string to produce one loop of a standing wave? (in kg)

B) What mass must be hung from this end of the string to produce two loops of a standing wave?
(in kg)

C) What mass must be hung from this end of the string to produce five loops of a standing wave?
(in kg)

Explanation / Answer

A) For one loop

fundamental frequency f = [1/(2L)]*v
here v is speed of the wave v = sqrt(T/mu)

Tesion T = v^2*mu

mu is the linear density

here T = m*g
then m*g = v^2*mu

but v = lamda*f =

for single loop

L = lamda/2
lamda = 2*L = 2*1.58 = 3.16 m
frequency f = 62

v = 3.16*62 = 195.92 m/s

then m*g = 195.92^2*4.34*10^-4

mass m = 1.69 Kg

B) For two loops
lamda = L = 1.58 m
v = Lamda*f = 1.58*62 = 97.96 m/s

But m*g= v^2*mu = (97.96)^2*(4.34*10^-4) = 4.16

mass = 4.16/9.81 = 0.424 kg

3) For five oops
L = 5*lamda/2

lamda = 2L/5 = (2*1.58)/5 = 0.632 m
v = 0.632*62 = 39.184 m/s

but m*g = v^2*mu = (39.184)^2*4.34*10^-4 = 0.6663

mass = m = 0.0679 kg

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