Question

• Problem 1:A particle moves in simple harmonic
motion about x = 0 with period
• 0.4 s. At t = 0, it has its maximum positive
acceleration, 28 m/s2
• (a) Find the amplitude and the phase constant
I of the oscillation.
(b) Write the displacement as a function of time
in the form x(t) = A sin( o. t + 0)​

Answer 1

Answer:

maximum

Explanation:

Given :

Time of period T = 0.4 s

max a = 28 m/s²

Angular velocity ω = 2π/T

= 2 x π /0.4

= 5π rad/s

a) acceleration amax = - A ω²

Since a is positive

amax = A ω²

28 = A (5π)²

Amplitude A = 28/25π²

= 28/[(25x (22/7)²]

= 28 x 7x 7/(25 x 22 x22)

= 0.113 m

Since at time t= 0 the particle is having maximum positive accln , it is crossing the mean position.

∴ phase constant Φ = 0

b) x(t) = A sin(ωt + Φ)

= 0.113 sin(5πt + 0)