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Question
When a mass m is connected individually to two springs S1 and S2, the oscillation frequencies are ν1 and ν2. If the same mass is attached to the two springs as shown in figure, the oscillation frequency would be ______.
Options
`v_1 + v_2`.
`sqrt(v_1^2 + v_2^2)`.
`(1/v_1 + 1/v_2)^-1`.
`sqrt(v_1^2 - v_2^2)`.
Solution
When a mass m is connected individually to two springs S1 and S2, the oscillation frequencies are ν1 and ν2. If the same mass is attached to the two springs as shown in figure, the oscillation frequency would be `underline(sqrt(v_1^2 + v_2^2))`.
Explanation:
When the mass is connected to the two springs individually,
`v_1 = 1/(2pi) sqrt(k_1/m)` ......(i)
`v_2 = 1/(2pi) sqrt(k_2/m)` ......(ii)
Now, the block is connected with two springs considered parallel.
Here equivalent spring constant, `k_(eq) = k_1 + k_2`
The time period of oscillation of the spring block system is
`T = 2pi sqrt(m/k_(eq))`
= `2pi sqrt(m/(k_1 + k_2))`
Hence frequency,
`v = 1/T = 1/(2pi) xx sqrt((k_1 + k_2)/m)` ......(iii)
`v = 1/(2pi) [k_1/m + k_2/m]^(1/2)`
From equation (i) `k_1/m = 4pi^2v_1^2` and from equation (ii), `k_2/m = 4pi^2v_2^2`
⇒ `v = 1/(2pi) [(4pi^2v_1^2)/1 + (4pi^2v_2^2)/1]^(1/2) = (2pi)/(2pi) [v_1^2 + v_2^2]^(1/2)`
⇒ `v = sqrt(v_1^2 + v_2^2`
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