SHMHard
Question
For a body executing SHM with amplitude A, time period T, maximum velocity vmax and phase constant zero, which of the following statements are correct for 0 ≤ t ≤ T/4 (y is displacement from mean position) ?
Options
A.At y = (A/2), v > (vmax/2)
B.for v = (vmax/2), y > (A/2)
C.For t = (T/8), y > (A/2)
D.For y = (A/2), t < (T/8)
Solution
For S.H.M
y = A sin ωt = A sin
v = ω
vmax = ωA
(A) At y = A/2
v = ω
= 
(B) For v =
⇒ ω
⇒ A2 − y2 =
⇒ y = 
(C) for t = T/8
y = A sin
(D) for y = A/2 A/2 = A sin
⇒
= π/6 ⇒ t = 
< T/8
y = A sin ωt = A sin
v = ω
vmax = ωA(A) At y = A/2
v = ω
= (B) For v =
⇒ ω⇒ A2 − y2 =
⇒ y = (C) for t = T/8
y = A sin
(D) for y = A/2 A/2 = A sin
⇒
= π/6 ⇒ t = < T/8Create a free account to view solution
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