A wire is 1.2 m long and has a mass of 0.027 kg. The wire is stretched between t

Question

A wire is 1.2 m long and has a mass of 0.027 kg. The wire is stretched between two points and when driven by an oscillator at 540 Hz you notice that the wire is oscillating with four antinodes (four loops) between the rigid supports
(a) What is the fundamental frequency of the wire?
Hz

(b) What is the fundamental wavelength of the wire?
m

(c) What is the velocity of the wave in the wire?
m/s

(d) What is the tension in the wire?
N

(e) What is the next lowest frequency (closest to, but lower than 540 Hz) that will produce a standing wave on the wire?
Hz

(f) How many antinodes does this standing wave have?

Explanation / Answer

when a wire is held at two points, then the standing waves are formed with the wavelengths satisfying the following relation

(L=nlambda/2)

where n is a natural number. And is the nth mode and is also equal to number of antinodes formed

This can be rewritten in terms of frequeny by using ( u = c/lambda)

where c is speed of wave in the wire.

( u = nc/2L)

(a) since for n=4 (4 antinodes), frequency is 540 Hz, thus

(540=4c/2L)

thus, (c/L = 270 Hz)

also, the fundamental frequency is given by n=1, thus ( u_0=c/2L = 135 Hz)

(b) The fundamental wavelength is at n=1, thus, (L=lambda_0/2)

thus, (lambda_0=2L=2*1.2=2.4 m)

(c) the velocity is same at all frequencies, in particular for the fundamental mode

thus, the velocity of wave will be (c= u_0lambda_0=135*2.4=324 m/s)

(d) the formula for the velocity is (c=sqrt{rac{T}{mu}})

where is mass per unit length for the string and T is tension.

Now, (mu = m/L = 0.027/1.2 = 0.0225 kg/m)

thus, (T=c^2mu = 324^2*0.0225=2362 N)

(e) the normal modes for frequency are given as earlier by the formula

( u=nc/2L = n u_0)

for 540 Hz we had n=4, thus for the frequency closest to this, we should have n=3, thus,

( u=3*135 = 405 Hz)

(d) NUmber of antinodes is given by n, thus number of antinodes is 3.

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