# In a stretched string the number of loops changes from 5 to 4 by adding weight of 0.018 kg inn sonometer experiment.the initial tension in the string is ----------kg weight. a)0.032 b)0.004 c)0.8 d)0.64.....plz explain with clear solution.......

1
by sweetygannerla

2015-11-08T18:33:05+05:30

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Answer is :  Mass/weight  tied to the string = 0.032 kg.   So tension is 0.032 kg weight.    If we do rough and quick calculation using calculus, we get 0.045 kg weight.

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A stretched string oscillates in mechanical standing waves.  Since on SOnometer the string is tightly fixed and not open, the length L is divided into an integral number of  loops = n.  The two ends of the string are nodes.  Each loop is λ/2  long.

L = n λ/2  =>  λ = 2 L / n        where  n = number of loops = 1 , 2, 3, 4, ....

Let the mass attached to the end of the string be M kg.  The tension T = M g.   Then μ = linear mass density of the string.

Velocity of  a wave = v = λ * f  = wave length * frequency
Velocity of transverse waves on the string = v
v = √(T/μ)
=> M = v² μ / g  =  (f² μ / g) λ²
∴  M = [ 4 f² L² μ / g ] / n²
∴ M  = K / n²      for  K = a constant.

M = K / 5²
M + 0.018 = K / 4²
=>  0.018 = K [1/16 - 1/25]        =>  K = 0.8 kg

So  M = 0.8 /5²  =  0.032 Kg

So the tension is 0.032 kg weight.
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