A light string with a mass per unit length of 2.70 g/m has its ends tied to two

Question

A light string with a mass per unit length of 2.70 g/m has its ends tied to two walls separated by a distance equal to three-fourths the length of the string (see the figure below). An object of mass m is suspended from the center of the string, putting a tension in the string.

A light string with a mass per unit length of 2.70 g/m has its ends tied to two walls separated by a distance equal to three-fourths the length of the string (see the figure below). An object of mass m is suspended from the center of the string, putting a tension in the string. Find an expression for the transverse wave speed in the string as a function of the mass of the hanging object. (Use the following as necessary: m.) v = What should be the mass of the object suspended from the string if the wave speed is to be 64.0 m/s?

Explanation / Answer

The angle a between the wall & string:
cos a = 3/4
sin a = sqrt (1 - 9/16)
sin a = [sqrt(7)]/4

Let T be the string tension.
Component of force acting upwards is = T Sin a
And for both strings it is = 2 T Sin a
And weight mg is acting downwards.

So,
2 T sin a = mg ----- (1)

(a)
The mass per unit length k = 0.0027 kg/m
V = sqrt ( T / k) ---- (2)
Since from (1) & (2)

V = sqrt ( mg / [2k sin a])

(b)
V^2 = mg / (2k sqrt(7)/4)
V^2 = 2 mg / k sqrt(7)
m = k sqrt(7) V^2 / 2g
m = sqrt(7) (0.0027 kg/m) (64 m/s)^2 / (2 * 9.8 m/s^2)

m = 1.49 kg

m = 1.5 kg (approx)

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