In the arrangement shown below, an object can be hung from a string (with linear

Question

In the arrangement shown below, an object can be hung from a string (with linear mass density = 0.002 00 kg/m) that passes over a light pulley. The string is connected to a vibrator (of constant frequency f), and the length of the string between point P and the pulley is L = 1.70 m. When the mass m of the object is either 9.0 kg or 16.0 kg, standing waves are observed; no standing waves are observed with any mass between these values, however.

(a) What is the frequency of the vibrator? (Note: The greater the tension in the string, the smaller the number of nodes in the standing wave.)

(b) What is the largest object mass for which standing waves could be observed?

Explanation / Answer

standing waves occurs for the wavelength of

lambda)n = 2L/n, where n = 1, 2, 3, ...........

lambda)n*f = sqrt(T/mu)

for hanging mass

T = mg

using these

2*L*f/n = sqrt(mg/mu)

m1 = 9 kg

m2 = 16 kg

L = 1.7 m

mass density = mu = 0.002 kg/m

if n is number of nodes for 9 kg then

number of nodes for 16 kg will be = n - 1

Now we have 2 equations

2*L*f/n = sqrt(m1*g/mu)

2*L*f/(n -1) = sqrt(m2*g/mu)

dividing these

(n - 1)/n = sqrt(m1/m2)

[(n - 1)/n]^2 = 9/16

(n - 1)/n = 3/4

1 - 1/n = 3/4

1/n = 1/4

n = 4

using this value of n

2*L*f/n = sqrt(m1*g/mu)

f = [4*sqrt(9*9.81/0.002)]/(2*1.7) = 247.18 Hz

B.

The largest mass will give a standing wave of 1 loop

n = 1

2*L*f/n = sqrt(Mmax*g/mu)

Mmax = 4*L^2*f^2*mu/g =

Mmax = 4*1.7^2*247.18^2*0.002/9.81 = 143.99 kg = 144 kg

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