 #1
Mahfuz_Saim
 17
 2
 Homework Statement:

In the figure below, AB= 2.5 m and AC= 8 m. The spring is stretched by 0.2 m and then released from rest at t=0 sec so that the spring executes SHM. At the time a pebble is thrown with a velocity V= 11.2 m/s from point C so that the pebble meets the block B at t= 1 sec. The angular frequency of the spring is π/3.
Question 1: Find the velocity of the spring at 1 sec.
Question 2: Will the pebble meet with block B according to the given condition?
 Relevant Equations:

v= Aω*cos(ωt+δ)
v= ω√(A^2x^2)
t=x/ v cosθ
I have used v= Aω*cos(ωt+δ) where A= 0.2 m, ω= π/3, t=1 and δ=0. Are the values right in this case? I am confused.
Question 2:
From question 1 I have got the value of V which is 9 m/s. By using v= ω√(A^2x^2), I have got the value of x. Now, do I need to add it with 2.5(distance between AB) or subtract from 2.5? I am confused. Please explain with the cause.
By using t=x/ v cosθ, I have got the value of x after t second of the projectile. By doing some simple calculation, I can get its distance from A. But there's no information about the hight of the block. So can I say they will meet if the distance from A of them are the same?
I hope you have understood my question properly. Please help me. Thank you.