Question 62: Waves

Question

A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating
in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the
wire is 50 N and the speed of sound is 320 ms−1, the mass of the string is

Accepted Answer

The fundamental mode in a pipe closed at one end and the second harmonic in a string are shown in the figure.
Fundamental frequency of a hollow pipe closed at one end is
$${\upsilon _p} = {v \over {4L}}$$
where, v = speed of sound, L = length of a pipe
Frequency of second harmonic of a string is
$${\upsilon _s} = {2 \over {2l}}\sqrt {{T \over \mu }}  = {1 \over l}\sqrt {{T \over \mu }} $$
where, T = tension of the string, m = mass per unit length of the string, l = length of the string
$${\upsilon _s} = {1 \over l}\sqrt {{T \over {{m \over l}}}}  = {1 \over l}\sqrt {{{Tl} \over m}}  = \sqrt {{T \over {ml}}} $$ ...... (i)
where m is the mass of the string
According to given problem, $${\upsilon _p} = {\upsilon _s}$$
$$	herefore$$ $${v \over {4L}} = \sqrt {{T \over {ml}}} $$ or $$m = {{16{L^2}T} \over {{v^2}l}}$$
Substituting the given values, we get
$$m = {{16 	imes {{(0.8)}^2} 	imes (50)} \over {{{(320)}^2} 	imes 0.5}} = 0.01$$ kg = 10 gram

Answer

5 grams

Answer

10 grams

Answer

20 grams

Answer

40 grams

Waves

A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ms−1, the mass of the string is

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