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Derive an expression Jor the time period (T) of a simple pendulum which may depend upon the mass (m) ofthe bob, length (l) of the pendulumand acceleration due to gravity (g) . |
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Answer» Solution :Let T = `km^(3) l^(@) G^(c )` where k is a DIMENSIONLESS constant. Writing the equation in dimensional form, we have `[M^(@)L^(@)T^(1)]= [M]^(a)[L]^(b)[LT^(-2)]^(c )=[M^(a)L^(b+c)T^(-2c)]` Equating exponents of M, L and T on both SIDES, we GET a=0 , b+c=0 , -2c =1. Solving the eq., we get a = 0 , b=1/2 c= -1/2 Hence , T=k`SQRT((l)/(g))`, where k is constant. |
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