SHMHard
Question
A particle executes S.H.M along a straight line with mean position x = 0. period 20 s and amplitude 5 cm. The shortest time taken by the particle to go from x = 4 cm to x = - 3 cm is
Options
A.4 s
B.7 s
C.5 s
D.6 s
Solution
Minimum phase difference between two position
= 53o + 37o = 90o
⇒ Time taken =
= 5s
OR
ω =
Form figure θ1 + θ2 = 53o + 37o = 90o or
θ = ωt ⇒
t ⇒ t = 5 sec
= 53o + 37o = 90o
⇒ Time taken =
= 5s OR
ω =
Form figure θ1 + θ2 = 53o + 37o = 90o or
θ = ωt ⇒
t ⇒ t = 5 secCreate a free account to view solution
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