1ans should be a)
A body executes simple harmonic motion under the action of a force F1
with a time period 4/5s. If the force is changed to F2 it executes S.H.M with time period 3/5 s. 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
b. 24/25
c. 35/24
d. 15/12
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7 Answers
49is the answer (d)...
actually i am totally unsure,,,i was trying to do this problem unconventionally using graph and thus could shortlist two options C and d....but more chances of correctness goes to d...[1][1][1]
1F1=ω2x
= (4∩2/T2)x
write
F1=ω2x
= (4∩2/T12)x
F1=ω2x
= (4∩2/T22)x
add RHS and LHS
it will be
F1 + F2=
= 4∩2/Tnew2x
find Tnew