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Two springs are joined and connected to ...

Two springs are joined and connected to mass m as shown. If the springs separately have force constants `K_(1)` and `K_(2)` , then frequency of oscillation of mass m is

A

`1/(2pi) sqrt((K_(1)K_(2))/((K_(1) + K_(2))m)`

B

`2pi sqrt((K_(1) + K_(2))m)/(K_(1)K_(2))`

C

`2pi sqrt((K_(1)K_(2))/m)`

D

`1/(2pi) sqrt(m/(K_(1) + K_(2)))`

Text Solution

Verified by Experts

The correct Answer is:
A
Let the body be gieven a displacement d.
Now, `d=-F//K`
If `d_(1)` and `d_(2)` be the extension in the two springs, then
`d_(1) =-F//K_(1)` and `d_(2) =-F//K_(2)`
But `d=d_(1) + d_(2)`
`therefore -F/K =-F/K_(1) -F/K_(2)` or `1/K =1/K_(1) + 1/K_(2)`
`therefore K = (K_(1)K_(2))/(K_(1) + K_(2))`
`n =1/(2pi) sqrt((K_(1)K_(2))/((K_(1) + K_(2))m)`
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Knowledge Check

  • Two spring of force constants K and 2K are connected a mass m below The frequency of oscillation the mass is

    A
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