What is the maximum acceleration when phase angle is given

							
a 
max
​ 
 =Aω 
2
      .....(1)
V 
max
​ 
 =Aω
Now, 
V 
max
​ 
 
a 
max
​ 
 
​ 
 = 
Aω
Aω 
2
 
​ 
 =ω
i.e =ω=10 
o
 
displacement =A 
m
​ 
 cos(ωt+θ)
=Acos(0+ 
4
1
​ 
 )
=Acos 
4
1
​ 
 
5=A 
2
​ 
 
1
​ 
 
⇒A=5 
2
​ 
 
Now a 
max
​ 
 =5 
2
​ 
 ×10 
2
  from (1)
=10×5 
2
​ 
 
=500 
2
​ 
 m/s 
2
						

							
Dear student 
Above ans is not  understandable 
Phase difference between acceleration and velocity in SHM is 90 degrees.
Let's see by practical example instead of trigonometry.
Case of swinging pendulum. When it's at mean position , acceleration is 0 because no restoring force acts on it. However velocity is maximum here. Now when pendulum reaches right extremity it's velocity becomes 0 at the moment of reversal, but here acceleration becomes maximum!
Now again when pendulum reaches mean position from right side it's velocity becomes maximum. But now again acceleration becomes 0.
This is quarter part of oscillation from mean to one extremity. Or from extremity to mean.
To arrive at maximum acceleration from maximum velocity pendulum has to travel 1/4 th part of oscillation means quarter part of 360 degrees. It's 90 degrees or Pi ÷2 radian. Same for its travel from minimum (0) acceleration to minimum (0) velocity.
This can also be proved by maths using equation of displacement in SHM taking
Displacement x=A sin wt considering travel from mean position. Differentiate for velocity and again for acceleration. You find that phase difference is 90 degrees.
						

							
Don't be so much confused due to irritable answers . 
max. Acceleration = AW²
Max. Velocity = AW
Their ratio as given will be W = 10 s-¹
Now the equation of SHM will be
, x= Asin(Wt + π/4)
At t=0 , x= 5 so, amplitude will 5/sin(π/4)
A=5√2
Max. Acceleration = AW²= 5√2 * 10²
= 500√2
Good regards thanks