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In (Q. 24.) the tension in string is T and the linear mass density of string is `mu`. The ratio of magnitude of maximum velocity of particle and the magnitude of maximum acceleration isA. `(1)/(2pi)sqrt((mul^(2))/(T))`B. `2pisqrt((mul^(2))/(T))`C. `(1)/(2pi)sqrt((T)/(mul^(2)))`D. `(1)/(4pi)sqrt((mul^(2))/(T))` |
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Answer» Correct Answer - A Here, `v_("max")=Aomega=2aomega` (at antinode A = 2a) and acceleration is `f_("max") = omega^(2)A=2aomega^(2)` `therefore" Ratio "=(2a omega)/(2a omega^(2))=(1)/(omega)` `because` Frequency, `v = (n)/(2l) sqrt(((T)/(mu))) = (2)/(2l) sqrt(((T)/(mu)))` (for second harmonic, n = 2) `=sqrt(((T)/(mu l^(2))))` `because " "omega=2piv` `therefore" Ratio "=(1)/(2piv)` `= (1)/(2pi) sqrt(((mu l^(2))/(T)))` |
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