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Question

A box executes simple harmonic motion under the action of a force F1 with a time period 45 s. If the force is changed to F2, it executes S.H.M with time period 35 s. If both the forces F1 and F2 act simultaneously in the same direction on the body, its time period in seconds will be

A
1225
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B
2425
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C
3524
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D
1512
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Solution

The correct option is A 1225
Let the body be displaced by a distance x. If the restoring force is F1, then the angular frequency of the resulting simple harmonic motion is given by
ω21=Km=Kxmx=F1mx .....(i)
where m is the mass of the body.
For force F2, we have, ω22=F2mx ..... (ii)
Now, ω2=F1+F2mx ..... (iii)
From (i), (ii) and (iii), we get
ω2=ω21+ω22
(2πT)2=(2πT1)2+(2πT2)2
1T2=1T21+1T22=1(45)2+1(35)2
1T2=2516+259=25×2516×9=25×25144
T=14425×25=1225s.
Hence, the correct choice is (a).

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