A particle of masses M = 4.2 kg is connected by strings between two vertical wal

Question

A particle of masses M = 4.2 kg is connected by strings between two vertical walls as shown in the figure below. The strings to the left of the particle makes an angle of theta = 35 degree with the wall on the left and the string to the right of the particle is uniform and horizontal of length L = 3.0 and having a small mass of m = 2 The particle bangs in equilibrium with the strings on the right resulting as shown in the figure above. Determine the frequency of the resulting string. Assume that = 9.8 m/s2.

Explanation / Answer

First we must find the Tension in the horizontal string.

Looking at the mass, we can find the sum of the forces keeping it in equilibrium.

The forces in the y direction are a component of Angled string Tension and its weight such that

FTcos 35 = (mg)

FT = (4.2)(9.8)/(cos 35)

FT = 50.25 N

Then in the x direction, the component of the angled strings tension must equal the tension of the horizontal string such that

FTsin35 = T

T = (50.25)(sin 35) = 28.8 N

From here, we can determine the velocity of waves on that string using

v = (F/) where F is the string tension and is the mass per unit length of the string

= (0.002 kg)/3 m = 6.67 X 10-4 kg/m

So v = (28.8)/(6.67 X 10-4)

v = 207.8 m/s

Finally we know that v = f

You can see in the picture that there are 7.5 waves on the resonating string. The wavelenth is therefore 3/7.5 = .4m

f = v/

f = (207.8)/(.4)

f = 519.5 Hz

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