0 votes
280 views
in Chapter5:Oscillations by (3.2k points)
edited

Using differential equation of linear S.H.M, obtain the expression for (a) velocity in S.H.M., (b) acceleration in S.H.M.

2 Answers

0 votes
by (3.2k points)
selected by
 
Best answer

The general expression for the displacement of a particle in SHM at time t is

x = A sin(ωt + α) ….. (1)

where A and ω is a constant in a particular case and α is the initial phase.

The velocity of the particle is 

\(v=\frac{dx}{dt}=\frac{d}{dt}[A sin(ωt + α)]\)

= ωA cos(ωt + α)

= \(ωA\sqrt{1-sin^2(ωt + α)}\)

From Eq. (1), sin(ωt + α) = x/A

\(v=ωA\sqrt{1-\frac{x^2}{A^2}}\)

∴ \(v=ω\sqrt{A^2-x^2}\)  …… (2)

Equation (2) gives the velocity as a function of x

The acceleration of the particle is

a= \(\frac{dv}{dt}=\frac{d}{dt}\)[Aω cos(ωt + α)]

∴ a = − ω2A sin(ωt + α)

But from Eq. (1),  A sin(ωt + α) = x

∴ a = − ω2x  ……. (3)

Equation (3) gives the acceleration as a function of x. The minus sign shows that the direction of the acceleration is opposite to that of the displacement.

0 votes
by (3.2k points)
edited
Video Explaination :: Youtube

Related questions

0 votes
1 answer 161 views
0 votes
1 answer 157 views

Doubtly is an online community for engineering students, offering:

  • Free viva questions PDFs
  • Previous year question papers (PYQs)
  • Academic doubt solutions
  • Expert-guided solutions

Get the pro version for free by logging in!

5.7k questions

5.1k answers

108 comments

557 users

...