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does an position in SHM is always a point of stable equilibrium ?

does an position in SHM is always a point of stable equilibrium ?

Grade:12th pass

3 Answers

Kenny Toijam
56 Points
7 years ago
First coming to the equilibrium part-For the particle under SHM to be in equilibrium it ks obvious that Force acting on the particle must be 0 that means equation for SHM is a=-bx(b=omega squared) so putting a=0(since m cannot be 0 for F to be 0) we have x=0 that means the body is at dynamic quilibrium at its equilibrium (mean)position only.As the body passes through the mean position and start moving towards lets suppose +ve x direction then F acts in opposite direction -ve x so this makes the equilibrium stable.
Saurabh Sharma
24 Points
7 years ago
IN SHM , BY DEFINITION 
ACCELARATION SHOULD ALWAYS BE DIRECTED TOWARD CENTRE
CONSIDER A POINT ,
BEFORE AND AFTER ACROSSING THAT POINT ACC. WILL ALWAYS BE DIRECTED TOWARD A PARTICULAR DIRECTION
Ankit Jaiswal
165 Points
6 years ago
Hi pranav, First coming to the equilibrium part-For the particle under SHM to be in equilibrium it ks obvious that Force acting on the particle must be 0 that means equation for SHM is a=-bx(b=omega squared) so putting a=0(since m cannot be 0 for F to be 0) we have x=0 that means the body is at dynamic quilibrium at its equilibrium (mean)position only.As the body passes through the mean position and start moving towards lets suppose +ve x direction then F acts in opposite direction -ve x so this makes the equilibrium stable.
 

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