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In the figure, a string, tied to a sinusoidal oscillator at P and running over a

ID: 1327769 • Letter: I

Question

In the figure, a string, tied to a sinusoidal oscillator at P and running over a support at Q, is stretched by a block of mass m. Separation L = 1.4 m, linear density ? = 1.9 g/m, and the oscillator frequency f = 190 Hz. The amplitude of the motion at P is small enough for that point to be considered a node. A node also exists at Q.

(a) What mass m allows the oscillator to set up the fourth harmonic on the string?
(b) What standing wave mode, if any, can be set up if m = 4 kg (Give 0 if the mass cannot set up a standing wave)?

Oscillator p

Explanation / Answer

a) speed of standing wave=v=sqrt(tension/mass per unit length)

here tension in the string=weight of the block=m*g=m*9.8

mass per unit length=linear mass density=1.9 g/m=0.0019 kg/m

to set up fourth harmonic,

the length of the string will be equal to 2*wavelength of the wave.


hence wavelength=L/2=0.7 m

as we know,

wavelength*frequency=speed of the wave


==>0.7*190=sqrt(m*9.8/0.0019)

==> m=3.4295 kg


hence 3.4295 kg mass will allow to set up fourth harmonic.


part b:

with mass m=4 kg, speed of wave=sqrt(m*g/0.0019)

=143.6369 m/s

as we know, wavelength*frequency=speed


let n th harmonic is set up in the string


then length of the string=(n/2)*wavelength

==> wavelength=(2/n)*1.4=2.8/n m

using wavelength*frequency=speed

we get

(2.8/n)*190=143.6369

==> n=3.7

hence no standing wave can be set up with m=4 kg.

(as value of n has to be an even integer to set up a standing wave)