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A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k|x|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.A. proportional to `1//sqrt(a)`B. independent of `a`C. proportional to `sqrt(a)`D. proportional to `a^(3//2)` |
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Answer» Correct Answer - A `V (x) = k|x|^(3)` `:. [k] ([V])/([x]) = (ML^(2)T^(2))/(L^(3)) = ML^(-1)T^(-2)` Now time period on `T` and `("mass")^(x) ("amplitude")^(y) (k)^(z)` `[M^@L^@T]=[M]^(x)[L]^(y)[ML^(-1)T^(-2)]^(z)` =`[M^(x+y)L^(y-x)T^(-2z)]` Equating the powers, we get `-2z =1` or `z = -1//2` `y-z = 0 y = z = - 1//2` Hence `T prop ("amplitude")^(-1//2)` or `T prop (1)/(sqrt(a))`. |
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