A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t = 0 it is at position x = 5 cm from mean position and going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t = 4s.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(x = (10 cm) sin 
, 
≈ 11 cm/s 2 )
A = 10 cm, T = 6 sec
So x = A sin 
= (10 cm) sin 
= 10 cm sin 
at t = 0, x = 5 cm
⇒ 5 cm = 10 cm sin (0 + ɸ)
⇒ 
= sin (ɸ)ɸ = 
So, x = (10 cm) sin 
acceleration a = – ꞷ
2 (10 cm) sin 
= – 
× (10 cm) sin 
= 
~ 11 cm/s 2
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems