Question

4. The velocity v of a wave in a stretched string depends on the tension T in the spring and the mass per unit length mu of the string. Obtain an expression for v in terms of T and mu, using the method of dimensions.

Accepted Answer

Given: v is proportional to T^x mu^y, where T=Tension in the spring, mu=mass per unit length, v=velocity of a wave. Thus v=kT^x mu^y, where k is a dimensionless constant. Dimensionally: T(N)=kgm/s^2, T(N)=MLT^-2; mu=kgm^-1, mu=ML^-1. Thus, LT^-1=k[MLT^-2]^x[ML^-1]^y. LT^-1=k[m^x L^x T^-2x][M^y L^-y]. LT^-1=kM^(x+y)L^(x-y)T^-2x. Comparing the L.H.S and R.H.S terms: For M: x+y=0 (1). For L: x-y=1 (2). And for T: -1=-2x, x=1/2, and from (1): x+y=0, y=-1/2. Thus, v=kT^(1/2)mu^(-1/2), or v=k*sqrt(T/mu).

WAEC Physics 2020 Theory — Question 4

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