InterviewSolution
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InterviewSolution
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Derive an expression for time period (t) of a simple penduleum, which may depend upon : mass of bob (m), length of pendulum (I) and acceleration due to gravity(g). |
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Answer» Let (i) `t prop m^(a) " "`(ii) `t prop l^(b) " "` (iii) `t prop g^(c)` Combining all the three factors, we get `t prop m ^(a)l^(b) g^(c) " "` or ` " " t=Km^(a)l^(b) g^(c) ` where K is a dimensionless constant of proportionality . Writing down the dimensions on either side of equation (i) , we get `[T] = [M^(a)] [L^(b)] [LT^(-2)]^(c) = [M^(a) L^(b+c)T^(-2c)]` Comparing dimensions, `a=0 ,b+c=0 ,-2c=1 ` `:. a=0, c=-1//2, b=1//2 ` From equation (i) `t=Km^(0)l^(1//2)g^(-1//2)) " " ` or ` " "t=K((l)/(g))^(1//2) =K sqrt((l)/(g))` |
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