Question

Particle is performing s.H.M. Along x-axis with amplitude 4 cm and time period 1.2 sec. The minimum time taken by the particle to move from x=+2cmx=+2cm to x=+4x=+4 cm and back again is given by

Answer 1

Hey dear,

◆ Answer-
T = 0.4 s

◆ Explaination-
# Given-
T = 1.2 s
A = 4 cm

# Solution-
Eqn of SHM starting from midpoint is-
x = Asin(ωt)

For x = +2 cm,
2 = 4sin(ωt)
sin(ωt) = 1/2
ωt = sininv(1/2)
2πt/T = π/6
t = T/12

For x = +4 cm,
4 = 4sin(ωt)
sin(ωt) = 1
ωt = sininv(1)
2πt/T = π/2
t = T/4

Time taken for x = 2 to x = 4 cm,
t = T/4 - T/12
t = T/6

Time taken to move back iz same as t. Therefore total time taken is
t' = 2t = T/3 = 1.2/3 = 0.4 s

Therefore total time taken is 0.4 s.

Hope this helps...

Answer 2

Answer:

0.4 s

see attachment may it will help you good luck