SHMHard
Question
A point particle of mass 0.1 kg is executing SHM of amplitude 0.1 m When the particle passes through the means position, its KE is 8 × 10-3 J. Find the equation of motion of this particle if the initial phase of oscillation is 45o.
Options
A.y = 0.1 cos (3t + π/4)
B.y = 0.1 sin (6t + π/4)
C.y = 0.1 sin (4t + π/4)
D.y = 0.1 cos (4t + π/4)
Solution
At mean position KE = 
m2ω2 = 8 × 10-3 ⇒ ω = 4 red /sec.
Now let equation of SHM be y = 0.1 sin (4t + φ). At t = 0, φ = 45o =
. Therefore y = 0.1 sin 
m2ω2 = 8 × 10-3 ⇒ ω = 4 red /sec.Now let equation of SHM be y = 0.1 sin (4t + φ). At t = 0, φ = 45o =
. Therefore y = 0.1 sin Create a free account to view solution
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