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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_2 = 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

Maximum linear momentum in case 1 is `(p_1)_(max) = mv_(max) b = m[aw_1]` …(i)
Maximum linear momentum in case 2 is `(p_2)_(max) = mv_(max) R = m[R omega_2]`
:. `1 = m omega_2` …(ii)
Dividing (i) & (ii) `(b)/(1) = (a omega_1)/(omega_2)`
:. `(omega_1)/(omega_2) = (b)/(a) = (1)/(n^2)` :. (B) is a correct option.
Also `E_1 = (1)/(2) m omega_1^2 a^2`
`E_2 = (1)/(2) m omega_2^2 R^2`
:. `(E_1)/(E_2) = (omega_1^2)/(omega_2^2) xx (a^2)/(R^2) xx (omega_1^2)/(omega_2^2) xx (1)/(n) = (omega_1^2)/(omega_2^2) xx (omega_2)/(omega_1) = (omega_1)/(omega_2)`
:. `(E_1)/(omega_1) = (E_2)/(omega_2)` (D) is the correct option.
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