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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

proportional to`1//sqrta`

B

independent of a

C

proportional tosqrta

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
`V(x)=K|x|^(3) therefore [k]=([V])/([x]^(3))=(ML^(2)T^(2))/(L^(3))=ML^(-1)T^(-2 )`
now time period T may depend on `(mass)^(x)(amplitude)^(y)(K)^(z)`
therefore `[M^(0)L^(0)T][M]^(x)[L]^(y)[ML^(-1)T^(2)]^(x)=[M^(x+y)L^(y-x)T^(-2x)]`
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)/(sqrta)`
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