A light string with a mass per unit length of 8.60 g/m has its ends tied to two

Question

A light string with a mass per unit length of 8.60 g/m has its ends tied to two walls separated by a distance equal to three-fourths the length of the string. An object of mass m is suspended from the center of the string, putting a tension in the string.

(a) Find an expression for the transverse wave speed in the string as a function of the mass of the hanging object. (Answer in terms of the mass m and note that this is the speed due to the mass/unit length and tension.)

(b) What should be the mass of the object suspended from the string in order to produce a wave speed of 45.0 m/s?
2 kg

Explanation / Answer

The angle q between the horizontal and either half of the string is such that:
cos q = 3/4
sin q = sqrt (1 - 9/16) = sqrt (7)/4

Let T be the string tension. Neglecting the mass of the string itself, the vertical projection of the tension of either half of the string is T sin q.
The sum of those two forces balances the vertical weight mg:
2 T sin q = mg
T is expressed in newtons (N) if m is in kilograms (kg) and g = 9.80665 m/s^2

(a) If k = 0.0086 kg/m is the mass per unit length, the celerity of the transverse wave is:
V = sqrt ( T / k) = sqrt ( mg / [2k sin q] )

(b)
V^2 = mg / (2k sqrt(7)/4) = 2 mg / k sqrt(7)
m = k sqrt(7) V^2 / 2g = sqrt(7) (0.0086 kg/m) (45m/s)^2 / (2 * 9.81 m/s^2)
m = 2.348... = 2.35 kg

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