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Two independent harmonic oscillators of ...

Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies `(omega_1) and (omega_2) and have total energies (E_1 and E_2), respectively. The variations of their momenta (p) with positions (x) are shown (s) is (are).

A

`E_(1)omega_(1)=E_(2)omega_(2)`

B

`(omega_(2))/(omega_(1))=n^(2)`

C

`omega_(1)omega_(2)=n^(2)`

D

`(E_(1))/(omega_(1))=(E_(2))/(omega_(2))`

Text Solution

Verified by Experts

The correct Answer is:
B, D

Refer of figure,
amplitude `A=a`
`p_(max)=b `or `m upsilon_(max)=b`
or `ma omega_(1)=b` ...(i)
`E_(1)=(1)/(2)ma^(2)omega_(1)^(2)` ...(ii)
Refer to figure
`p_(max)=R,` amplitude `A=R`
`m upsilon_(max)_(2))=R` or `m R omega_(2)=R` or `momega_(2)=1` ...(iii)
`E_(2)=(1)/(2)mR^(2)omega_(2)^(2)` ...(iv)
From (ii) and (iv),
`(E_(1))/(E_(2))=(a^(2)omega_(1)^(2))/(R^(2)omega_(2)^(2))=(n^(2)omega_(1)^(2))/(omega_(2)^(2))` ...(v)
`[:. (a)/(R)=n]`
From (i) and (iii)
`(aomega_(1))/(omega_(2))=(b)/(2)` or `(omega_(1))/(omega_(2))=(b)/(a)=(1)/(n^(2))` ..(vi)
or `(omega_(2))/(omega_(1))=n^(2)` `[`options (b) is true `]`
Putting the above value in (v) , we get
`(E_(1))/(E_(2))=(omega_(2))/(omega_(1))xx(omega_(1)^(2))/(omega_(2)^(2))=(omega_(1))/(omega_(2))` or `(E_(1))/(omega_(1))=(E_(2))/(omega_(2))`
Hence, option (d) is true.
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