Question

A uniform wire of length L and mass m is fixed between two points having tension F . A sound of frequency f is impressed on it . Then find the vibrational energy in terms of FLM.​

Answer 1

Answer:

Explanation:

The frequency of the wire (u) depends on length of string (l), tension(F) and mass per unit length of the wire (m= M/L).

Thus,

u α $L^{a}F^{b} m^{c}$- eq.1

Now lets write the above equation with their respective dimensions,

$M^{0}L^{0} T^{-1}$ = $L^{a}[M^{1} L^{1} T^{-2}]^{b}$ $[M^{0}L^{-1}]^{c}$

Multiply the dimensions in the above equation,

                    =$M^{b+c}L^{a+b-c} T^{-2b}$

Here, b + c = 0 ;  

         a + b − c = 0  and − 2b = − 1

On solving, we get

a = −1 , b = 1/2 and c = −1/2

Now substitute these values of a,b, and c in eq.1 and introduce a constant k,

u α $L^{-1}F^{1/2} m^{-1/2}$

i.e, u= k 1/L$\sqrt\frac{F}{m}$

We know that m = M/L, substitute this in the above equation,

u=  k 1/L$\sqrt\frac{F}{M/L}$

Thus, u = k $\sqrt\frac{F}{ML}$

The value of constant k is 1/2 and substitute it,

u = $\frac{1}{2}$$\sqrt\frac{F}{ML}$

Answer 2

Answer:

The frequency of wire depends on the length and the tension and also mass per unit length of the wire.

The vibrational energy can be calculated, and it comes to ½  X root F/ML.  

This is one of the major conditions of travelling of sound in motion and the distance travelled, and the tension is directly responsible for the frequency of the sound.