Two springs, of force constant k1 and k2, are connected to a mass m as shown. The frequency of oscillation of the mass is f. If both k1 and k2 are made four times their original values, the frequency of oscillation becomes
A
f/2
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B
f/4
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C
4f
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D
2f
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Solution
The correct option is D2f This system can be treated as two parallel spring system with spring constants k1 and k2. Now, frequency f=12π√k1+k2m When both k1 and k2 are made four times their original value, the new frequency becomes: f′=12π√4k1+4k2m =12π2√k1+k2m =2f