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Expression for derive an expression for displacement of a particle executing SHM

Answer 1

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Secondary School Physics 13 points

Derive the expressions for displacement, velocity and acceleration of a particle executes S.H.M.

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abhi178

abhi178 Genius

Let's consider a particle P moves on the circumference of circle of radius A with uniform acceleration \omega.Let PN is the perpendicular drawn from P to diameter yy'.

see figure, \angle{POX}=\theta, OP = A and ON = y

from triangle ∆ONP,

sin\theta=\frac{ON}{OP}

we know, \theta=\omega t

so, OPsin\omega t=ON

or, \boxed{y=Asin\omega t} ....(i),This is the expression of equation of SHM.

now differentiate y with respect to t, to find velocity of particle.

\frac{dy}{dt}=v=\omega Acos\omega t

or, \boxed{v=\omega Acos\omega t} .....(ii)

from equations (i) and (ii),

\boxed{\bf{v=\omega\sqrt{A^2-y^2}}} this is the expression of velocity of particle in SHM.

again , differentiate v with respect to t ,

Answer 2

Answer:

derive and expression for the displacement of a particle executing SHM