Please solve the problem
Asked by | 27th Oct, 2008, 08:11: PM
L=0.25m, m = 0.0025kg
The first overtone in the string has a frequency of n_string = 2.(1/2L square root of T/m). Note that this is twice the fundamental frequency.
The fundamental frequency of the pipe closed at one end n_pipe = v/4L
Here v = 330m/s and L = 0.4m which is the length of the closed pipe.
The frequency of the string will be n_pipe +8 because the question says that on decreasing the tension the beat frequency decreases. If the frequency of the string was n_pipe - 8 then on decreasing the tension the beat frequency would increase.
Once you know n_pipe, L of the string, mass per unit length = 0.0025kg/0.25m then the value of T can be found by subsituting the known values.
Answered by | 6th Dec, 2008, 01:05: PM
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