Question

The frequency of vibration of stretched string depends on its length its mass per unit length and the tension in the string obtain dimensionally an expression for frequency

Answer 1

Factors affecting frequency:

⇒ The frequency is directly proportional to the square root of tension.

⇒ The frequency is inversely proportional to the radius .

⇒ The frequency is inversely proportional to the density of the string.

The frequency can be increased by increasing tension , decreasing radius and density.

The frequency can be decreased by decreasing tension , increasing radius and density.

Expression :

F ∝ √T

F ∝ 1 / d

F ∝ 1 / r

Formula :

$\mathsf{F=\frac{1}{2l}\frac{\sqrt{T}}{r^2d}}$

Answer 2

The formula is

$\boxed{v=\frac{k}{l}\sqrt{\frac{T}{m}}}$

Explanation:

Let the frequency of vibration v depends upon, length l, tension T and mass per unit length m in the following way

$v=kl^aT^bm^c$, where k is a constant

Dimensions of v = [T⁻¹]

Dimensions of l = [L]

Dimensions of T = [MLT⁻²]

Dimensions of m = [ML⁻¹]

Thus,

LHS Dimensions =RHS Dimensions

$\implies [T^{-1}]=k[L]^a[MLT^{-2}]^b[ML^{-1}]^c$

$\implies  [T^{-1}]=k[M]^{b+c}[L]^{a+b-c}[T]^{-2b}$

Comparing the dimensions on both sides

We get

$b+c=0$   ............. (1)

$a+b-c=0$  ..............(2)

$-2b=-1$  ................... (3)

From eq (3)

$b=\frac{1}{2}$

Thus, from eq (1)

$c=-\frac{1}{2}$

And from eq (2)

$a+\frac{1}{2}-(-\frac{1}{2})=0$

$\implies a+1=0$

$\implies a=-1$

Thus, the equation becomes

$v=kl^{-1}T^{1/2}m^{-1/2}$

or, $v=\frac{k}{l}(\frac{T}{m})^{1/2}$

or, $v=\frac{k}{l}\sqrt{\frac{T}{m}}$

Thus is the required equation.

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