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Physics Question on Oscillations

The frequency of vibration ff of a mass mm suspended from a spring of spring constant kk is given by a relation f=amxkyf = am^x k^y, where aa is a dimensionless constant. The values of xx and yy are

A

x=12,y=12x=\frac{1}{2}, y=\frac{1}{2}

B

x=−12,y=−12x=-\frac{1}{2}, y=-\frac{1}{2}

C

x=12,y=−12x=\frac{1}{2}, y=-\frac{1}{2}

D

x=−12,y=12x=-\frac{1}{2}, y=\frac{1}{2}

Answer

x=−12,y=12x=-\frac{1}{2}, y=\frac{1}{2}

Explanation

Solution

f=amxky                   ....(i)f=am^x k^y \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ....(i) Dimensions of frequency f=[M0L0T−1]f=[M^0 L^0 T^{-1}] Dimensions of constant a=[M0L0T0]a=[M^0 L^0 T^0] Dimensions of mass m=[M]m = [M] Dimensions of spring constant k=[MT−2]k = [MT^{-2}] Putting these value in equation (i)(i), we get [M0L0T−1]=[M]x[MT−2]y[M^0 L^0 T^{-1}]=[M]^x[MT^{-2}]^y Applying principle of homogeneity of dimensions, we get x+y=0.....(ii)x+y=0 .....(ii) −2y=−1......(iii)-2y=-1 ......(iii) or y=12,x=−12y=\frac{1}{2}, x=-\frac{1}{2}