The frequency 'n' of vibration of stretched string depends on its length 'L' its mass per unit length 'm' and the tension 'T' in the string obtain dimensionally an expression for frequency 'n'.

### Asked by jullypradhan001 | 5th Aug, 2018, 11:48: PM

Expert Answer:

### Frequency is denoted by 'n'
let us consider
n= kl^{x}T^{y}m^{z} ...(1)
Writing dimension for each quantity,
n → [M^{0}L^{0}T ^{-1}]
l→ [M^{0}L^{1}T^{0}]
T→[M^{1}L^{1}T^{-2}]
m→[M^{1}L^{-1}T^{0}]
Substituting dimensions in equation (1),
[M^{0}L^{0}T^{-1}] = k[M^{0}L^{1}T^{0}]^{x} [M^{1}L^{1}T^{-2}]^{y} [M^{1}L^{-1}T^{0}]^{z}
[M^{0}L^{0}T^{-1}] = k [M^{y+z} L^{x+y-z} T^{-2y}] ...(2)
From (2),
y+z = 0
x+y-z =0
-2y = -1 ....(3)
Solving equations (3) simultanously we get,
x = -1, y = 1/2 and z = -1/2
Substituting these value of x,y and z in equation (1),

n= kl^{-1}T^{1/2}m^{-1/2}
This is the formula for frequency of vibration for stretched string.

^{x}T

^{y}m

^{z}...(1)

^{0}L

^{0}T

^{-1}]

^{0}L

^{1}T

^{0}]

^{1}L

^{1}T

^{-2}]

^{1}L

^{-1}T

^{0}]

[M

^{0}L^{0}T^{-1}] = k[M^{0}L^{1}T^{0}]^{x}[M^{1}L^{1}T^{-2}]^{y}[M^{1}L^{-1}T^{0}]^{z}^{0}L

^{0}T

^{-1}] = k [M

^{y+z}L

^{x+y-z}T

^{-2y}] ...(2)

^{-1}T

^{1/2}m

^{-1/2}

### Answered by Shiwani Sawant | 6th Aug, 2018, 12:18: PM

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