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Question

# A body executes simple harmonic motion under the action of a force F1 with a time period (4/5) seconds. If the force is changed to F2, it executes SHM with time period (3/5) seconds. If both the forces F1 and F2 act simultaneously in the same direction on the body, its time period (in seconds) is

A
12/25
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B
24/25
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C
35/24
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D
25/12
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Solution

## The correct option is A 12/25F1=−k1x,F2=−k2x F1=−(k1m)x;F2=−(k2m)x ⇒F1=−ω21x;f2=−ω22x Now, resultant force F=F1+F2=−k1x−k2x ⇒−kx=−k1x−k2x ⇒k=k1+k2⇒mω2=mω21+mω22 ⇒ω2=ω21+ω22 ⇒(2πT)2=(2πT1)2+(2πT2)2 ⇒1T2=1T21+1T22 ∴T=T1T2√T21+T22 =45×35√(45)2+(35)2=1225 second Hence, the correct answer is option (a).

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