Deepak Patra
Last Activity: 10 Years ago
Sol. x = (2.0cm)sin [(100s–1) t + (π/6)]
m = 10g.
a) Amplitude = 2cm.
ω = 100 sec–1
∴ T = 2π/100 = π/50 sec = 0.063 sec.
We know that T = 2π √(m/k) ⇒ T2 = 4π2 × m/k ⇒ k = 4 π2/T2 m
= 105 dyne/cm = 100 N/m. [because ω = 2π/T = 100 sec-1]
(b) At = 0
x = 2cm sin (π/6) = 2 × (1/2) = 1 cm. from the mean position.
We know that x = A sin (ωt + ∅)
v = A cos (ωt + ∅)
= 2 × 100 cos (0 + π/6) = 200 × √3/2 = 100 √3 sec-1 = 1.73m/s
c) a = – ω2 x = 1002 × 1 = 100 m/s2