A string of length L = 1.2 m is attached at one end to a wave oscillator, which

Question

A string of length L = 1.2 m is attached at one end to a wave oscillator, which is vibrating at a frequency L = 80 Hz. The other end of the string is attached to a mass hanging over a pulley as shown in the diagram below. When a particular hanging mass is suspended from the string, a standing wave with two segments is formed. When the weight is reduced by 2.2 kg. a standing wave with five segments is formed. What is the linear density of the string? 5) A standing wave is formed in a string as pictured in problem 1. If a standing wave with two segments is formed when the frequency is 60 Hz. and the hanging mass and string length are kept constant, what frequency would be required to produce a standing wave with 7 segments?

Explanation / Answer

4 Frequency kept constant & mass is varied

wavelenth=2l/n; l= lenth of string,n=no of segments.

velocity=wavelenth * Frequency = sqrt(T/u) ; u= linear mass density; T = tension in string; f= frequency

T=4ul2f2/n2

1) T= mg = 4ul2f2/n2 ; n=2

2) T= (m-2.2)g = 4ul2f2/n2 ; n=5

1/2 =

gives m=2.62 kg

Putting m in 1; we have u=.0028 kg/m ;Take g =10

5 f varied & T kept constant

1) T = 4ul2f2/n2 ; f=60 & n=2

2) T = 4ul2f2/n2 ; f=? & n=7

1/2= we get;

f = 210 Hz

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