Question

# If $$n_1, \space n_2, \space n_3, \space ...$$ are frequencies of segments of a stretched string, the frequency $$n$$ of the string is given by

A
n=n1×n2×n3×...
B
n=n1+n2+n3+...
C
1n=1n1+1n2+1n3+...
D
none of the above

Solution

## The correct option is B $$\displaystyle\frac{1}{n} = \displaystyle\frac{1}{n_1}+\displaystyle\frac{1}{n_2}+\displaystyle\frac{1}{n_3}+...$$Frequency of the first segment can be given by: $$n_1=\dfrac{1}{2l_1}\sqrt{\dfrac{T}{m}}\Rightarrow l_1=\dfrac{1}{2n_1}\sqrt{\dfrac{T}{m}}$$Similarly the frequency of the second segment can be given by: $$n_2=\dfrac{1}{2l_2}\sqrt{\dfrac{T}{m}}\Rightarrow l_2=\dfrac{1}{2n_2}\sqrt{\dfrac{T}{m}}$$ and so on. The total length of the string can be given by: $$l=l_1+l_2+....=\dfrac{1}{2n_1}\sqrt{\dfrac{T}{m}}+ \dfrac{1}{2n_2}\sqrt{\dfrac{T}{m}}+...=d\dfrac{1}{2}\sqrt{\dfrac{T}{m}}\left [\dfrac{1}{n_1}+\dfrac{1}{n_2}+...\right ]$$$$\Rightarrow 1=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}}\left [\dfrac{1}{n_1}+\dfrac{1}{n_2}+...\right ]=\nu\left [\dfrac{1}{n_1}+\dfrac{1}{n_2}+...\right ]$$$$\Rightarrow \dfrac{1}{n}=\left [\dfrac{1}{n_1}+\dfrac{1}{n_2}+...\right ]$$Physics

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