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Question: The frequency of vibration \(f\) of a mass \(m\) suspended from a spring of spring constant \(K\)is ...

The frequency of vibration ff of a mass mm suspended from a spring of spring constant KKis given by a relation of this type f=CmxKyf = Cm^{x}K^{y}; where CC is a dimensionless quantity. The value 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

By putting the dimensions of each quantity both the sides we get [T−1]=[M]x[MT−2]y\lbrack T^{- 1}\rbrack = \lbrack M\rbrack^{x}\lbrack MT^{- 2}\rbrack^{y}

Now comparing the dimensions of quantities in both sides we get x+y=0 6muand 2y=1x + y = 0\mspace{6mu}\text{and }2y = 1 ∴\therefore x=−12,y=12x = - \frac{1}{2},y = \frac{1}{2}