InterviewSolution
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InterviewSolution
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The frequency `f` of vibrations of a mass `m` suspended from a spring of spring constant `k` is given by `f = Cm^(x) k^(y)` , where `C` is a dimensionnless constant. The values of `x and y` are, respectively,A. `(1)/(2) , (1)/(2)`B. ` -(1)/(2) , -(1)/(2)`C. `(1)/(2) , - (1)/(2)`D. `- (1)/(2) , (1)/(2)` |
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Answer» Correct Answer - D `f = Cm^(x)k^(y)`. Writing dimensions on both sides. `[M^(0)L^(0)T^(-1)] = M^(x) [ML^(0)T^(-2)]^(y) = [M^(x+y) T^(-2 y)]` Comparing dimensions on both sides , we have ` 0 = x + y and -1 = -2 y rArr y = (1)/(2) , x = -(1)/(2)` Aliter. Remembering that the frequency of oscillation of loaded spring is ` f = (1)/( 2 pi) sqrt(k)/( m) = (1)/( 2 pi) (k)^(1//2) m^(-1//2)` which gives `x = -(1)/(2) and y = (1)/(2)` |
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