CBSE Class 11-science Answered

The frequency f 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 f.

Asked by tps.mjmdr | 08 Jul, 2018, 05:52: PM

Expert Answer

The speed of a wave depends on the properties of the medium and is assumed to be independent of frequency and wavelength.

Each element of the string pulls on its neighbours with a force given by tension T in the string. The stronger the tension,

the greater the force between neighboring elements and the more rapidly any disturbance will propagate down the string.

Thus the wave speed should increase with increasing tension.

On the other hand, the inertia of each element limits how effective the tension will be in accelerating that element to move the

wave along the string. Thus for the same tension, the wave speed will be smaller in strings having more massive elements.

The mass of each small element can be given in terms of the mass density μ (mass per unit length ), which for a uniform string

is equal to its mass divided by its length. On the basis of these general principles we therfore expect

begin mathsize 12px style v space equals space T to the power of a over mu to the power of b end style

................(1)

where a and b are exponents that must be determined from the analysis.

Let us perform dimensional analysis for eqn.(1),

begin mathsize 12px style open square brackets m to the power of 1 s to the power of negative 1 end exponent close square brackets space equals space open square brackets k g space m space s to the power of negative 2 end exponent close square brackets to the power of a over open square brackets k g space m to the power of negative 1 end exponent close square brackets to the power of b space equals space left square bracket k g right square bracket to the power of a minus b end exponent space open square brackets m close square brackets to the power of a plus b end exponent open square brackets s close square brackets to the power of negative 2 a end exponent end style

hence we have a-b = 0; a+b = 1 and a=(1/2) , hence we get a = b = (1/2)

hence we rewrite eqn.(1) as

begin mathsize 12px style v space equals space C square root of T over mu end root space.............. left parenthesis 2 right parenthesis end style

where C is a non-dimensional constant. By performing mechanical analysis using newton's law it is found that C = 1;

hence we arrive the final formula for wave velocity

begin mathsize 12px style v space equals space square root of T over mu end root end style

frequency n = (v/λ ), where λ is wavelength

Answered by Thiyagarajan K | 13 Jul, 2018, 11:45: AM

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