Question

A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm.

Answer 1

Hii dear,

◆ Answer-

◆ Explaination-
# Given-
A = 5 cm = 5×10^-2 m
T = 0.2 s

# Solution-
(a) At x = 5 cm
Velocity is
v = ω√(A^2-x^2)
v = (2π/T)√(A^2-x^2)
v = (2π/0.2) √(0.05^2-0.05^2)
v = 0

Acceleration is
a = -xω^2
a = -x(2π/T)^2
a = -5×10^-2(2π/0.2)^2
a = -5π^2 m/s
a = -49.35 m/s^2

(b) At x = 3 cm
Velocity is
v = ω√(A^2-x^2)
v = (2π/T)√(A^2-x^2)
v = (2π/0.2) √(0.05^2-0.03^2)
v = 0.4π^2 m/s
v = 3.95 m/s

Acceleration is
a = -xω^2
a = -x(2π/T)^2
a = -3×10^-2(2π/0.2)^2
a = -3π^2 m/s
a = -29.61 m/s^2

(c) At x = 0 cm
Velocity is
v = ω√(A^2-x^2)
v = (2π/T)√(A^2-x^2)
v = (2π/0.2) √(0.05^2-0^2)
v = 0.5π^2 m/s
v = 4.93 m/s

Acceleration is
a = -xω^2
a = -x(2π/T)^2
a = -0×10^-2(2π/0.2)^2
a = 0 m/s^2

Hope this is helpful...