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The frequency of vibration of string dep...

The frequency of vibration of string depends on the length L between the nodes, the tension F in the string and its mass per unit length m. Guess the expression for its frequency from dimensional analysis.

Text Solution

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Frequency, ` f=KL^aF^bM^c
where M=Mass/unit length
L=Length
F=Tension (Force)
Dimension of f=[T^-1]`, Dimension of right side
`L^a=[L^b], F^b=[MLT^-2]^b
M^a=[ML^-1]^c
[T^-1]=K[L]^a[MLT^-2]^b[ML^-1]^c
[M^0L^0T^-1]=KM^(b+c)L^(a+b-c)T^(-2b)
Equating the dimensions of both sides
we get `b+c=0...........i
-c+a+b=0................ii
-2b=-1`
Solving the equatiions
we get
`a=-1
b =1/2
and a=-1
b=1/2
and c=(-1/2)
frequency f=KL^-1F^(1/2)M^(-1/2)
=K/L.F^(1/2)M^((-1/2))ltbrgeK/Lsqrt((F/M))`
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Knowledge Check

  • The frequency of vibration of string is given by v= p/(2l) [F/m]^(1/2) Here p is number of segments in the string and l is the length. The dimensional formula for m will be

    A
    `[M^0 L T^(-1)]`
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    `[ML^(-1) T^(0)]`
    D
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  • The frequency of vibration of string is given by v=p/(2l)[F/m]^(1//2) . Here p is number of segments in the string and l is the length. The dimensional formula for m will be

    A
    `[M^(0)LT^(-1)]`
    B
    `[ML^(0)T^(-1)]`
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    `[ML^(-1)T^(0)]`
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  • The frequency (n) of vibration of a string is given as n = (1)/( 2 l) sqrt((T)/(m)) , where T is tension and l is the length of vibrating string , then the dimensional formula is

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    B
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    C
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