In S. H. M. path length is 4 cm and maximum
acceleration is 27t2 cm/s2. Time period of
motion
(а) 2ѕ b) 2s (c) 4s (d) 1/2s
plzzz solve this and also explain it
Answers
It is given that, in simple harmonic motion, path length is 4cm and maximum acceleration, a = 2π² cm/s².
we have to find time period of motion.
amplitude = path length/2 = 2 cm
acceleration, a = ω²A
⇒2π² = ω² (2cm)
⇒ω = π rad/sec
time period, T = 2π/ω = 2π/π = 2 sec
therefore, time period is 2 sec.
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amplitude = path length/2 = 2 cm
acceleration, a = ω²A
⇒2π² = ω² (2cm)
⇒ω = π rad/sec
time period, T = 2π/ω = 2π/π = 2 sec