Question

A tube 1m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.30m long and has a mass of 0.010kg. It is held fixed at both ends and vibrates in its fundamental mode. It sets the air column in the tube into vibration at its fundamental frequency by resonance. Find (a) the frequency of oscillation of the air column and (b) the tension in the wire.

Answer #1

length of the tube L =1.0m

fundamental frequency of the tube

f = v / 4L

where v = speed of sound = 340 m / s

f=340/4*1

plug the values we get f = 85 Hz

resonance is occured.So, funadamental frequency of the wire f' =f = 85 Hz

we know f ' = ( 1/ 2L ' ) √[ T / μ]

from this tension in the wire T = μ * ( 2L ' f ' ) ^2

where L ' = length of the wire = 0.30 m

μ =linear density = 10 g / 0.30 m

= ( 10 * 10 ^ -3 kg ) / 0.30 m

= 0.033 kg / m

plug the values we get

T=(85)^2(4)(0.30)^2(0.033)

T = 85.83 N

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