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A particle of mass (m) is executing osci...

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

(a)porprotional to `(1)/(sqrt(a))`

B

(b)independent of a

C

(c)proportional to `sqrt(a)`

D

(d)proportional to `a^(3//2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`U(x) = k|x|^(3)`
`:. [k] = (|U|)/(|x^(3)|) = ([ML^(2)T^(-2)])/([L^(3)]) = [ML^(-1)T^(-2)]`
Now, time period may depend on
`T prop ("mass")^(x)("amplitude")^(y) (k)^(z)`
`rArr [M^(0)L^(0)T] = [M]^(x)[L]^(y)[ML^(-1)T^(-2)]^(2) = [M^(x+2)L^(y-2)T^(-2x)]`
Equating powers, we get
`2z = 1` or `Z = -1//2`
`y -z = 0` or `y = z = -1//2`
Hence, `Tprop("amplitude")^(-1//2)`
`rArr T prop (a)^(-1//2) rArr T prop (1)/(sqrt(a))`
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