Question

the frequency of a wave in a stretched string depends on the length L, tension T and linear density m with dimension ML-1 . deduce the formula for f in terms of L , T and M using dimensional analysis​

Answer 1

Answer:

$f = \frac{1 }{L}\sqrt{\frac{T}{m} }$

Explanation:

Let the frequency ,

$f = L^{x}T^{y}m^{z}$

Now the unit of frequency is hertz = s⁻¹ = T⁻¹ where T is time, tension T = kgm/s² = MLT⁻¹ and linear density m = kg/m = ML⁻¹.

So,

$T^{-1} = L^{x}[MLT^{-2} ]^{y}[ML^{-1} ]^{z}$

collecting the like bases, we have

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

Equating powers on both sides, we have

x + y - z = 0  (1)

y + z = 0       (2)

-2y = -1         (3)

From (3), y = 1/2

From (2), z = -y = -1/2

Substituting y and z into (1), we have

x + y - z = 0

x + 1/2 - (-1/2) = 0

x + 1/2 + 1/2 = 0

x + 1 = 0

x = -1

So,

$f = L^{x}T^{y}m^{z}\\f = L^{-1}T^{\frac{1}{2} }m^{\frac{-1}{2} }$

$f = \frac{1 }{L}\sqrt{\frac{T}{m} }$