A string (not necessarily of the same type you used in lab) is stretched horizon

Question

A string (not necessarily of the same type you used in lab) is stretched horizontally, and one end is connected to the tip of a mechanical oscillator. The other end extends over a pulley and supports a hanging mass. The string is L = 3.35 meters long, from the point of attachment to the oscillator to the point of contact with the pulley. In order to determine the mass density of the string, you cut another piece of it from the spool, which was delta l = 2.6 m long, and found that its mass was delta m = 1.23 grams. When the frequency of the function generator is set to f = 25.5 Hz. a standing wave pattern as the one shown in the first figure is formed on the string, and when the frequency is set to f = 42.5 Hz. a standing wave pattern as the one shown in the second figure is formed on the string. (See the figures below: NOT ACTUAL SIZE). From the figure, determine the wave speed nu on this stretched string. Include units. What the linear mass density mu of the string. Include units. Determine the tension F_tension in the string. Include units.

Explanation / Answer

a) for first figure :

1.5 lambda = L

labda = 3.35/1.5 = 2.23 m

and v = lambda*f = 2.23 * 25.5 = 56.95 m/s

for second figure:

2.5 lambda = L

lambda = 3.35/2.5 = 1.34 m

v = lambda * f =1.34 * 42.5 = 56.95 m/s

so speed of wave is 56.95 m/s.

b) linear mass density = mass / length

         = 1.23 * 10^-3 kg / 2.6m = 4.73 x 10^-4 kg/m

c) Speed of string Wave, v = sqrt[T/linear mass density]

       = sqrt[56.95 / (4.73 x 10^-4)] = 346.96 N

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