Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 3 (Q.No. 31)
31.
The acceleration of a particle moving with simple harmonic motion is __________ at themean position.
Discussion:
8 comments Page 1 of 1.
SAURABH GANGWAR said:
9 years ago
Please explain it.
UMESH BAROT said:
9 years ago
At this position the displacement is maximum.
Jun Arro said:
8 years ago
Let's understand it as " the mean position".
So, the mean position is at the centre of the pendulum motion. Clearly, at the centre there is no motion and hence, acceleration is zero.
So, the mean position is at the centre of the pendulum motion. Clearly, at the centre there is no motion and hence, acceleration is zero.
Ravi p m said:
8 years ago
We know that velocity= differentation(s/t).
And acceleration = differentation(v/t).
since at mean position displacement(s)=0.
then velocity =0 and acc =0.
s=displacement.
v velocity.
And acceleration = differentation(v/t).
since at mean position displacement(s)=0.
then velocity =0 and acc =0.
s=displacement.
v velocity.
Abhishek said:
8 years ago
Acceleration= w^2 y.
y=0.
than acceleration becomes 0.
y=0.
than acceleration becomes 0.
Hriday said:
7 years ago
It is for instant acceleration. A= w^2 are where r is zero.
Kaushar Raza said:
6 years ago
Option A is the correct answer.
Kamran said:
4 years ago
Their is no restoring force acting while it is on mean position so acceleration is zero.
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