Physics, asked by Shawel, 1 year ago

QUESTION---The acceleration of a particle, given by relation,
a= -5w^2 sin(wt). At t = 0, X = 0 and v= 5w. The
displacement of this paticle at time t will be
(1) -5 sin(wt)
(2) 5 sin(wt)
(3) 5 cos(wt)
(4) -5 cos(wt)

plsss tell fast , test tomorrow!!!​

Answers

Answered by sweetysoya
40

Answer:

hey see the attached solution

Attachments:
Answered by CarliReifsteck
8

Given that,

The equation of acceleration is

a=-5\omega^2\sin(\omega t)

We need to calculate the displacement of this particle at time t

Using equation of acceleration

a=-5\omega^2\sin(\omega t)

\dfrac{dv}{dt}=-5\omega^2\sin(\omega t)

On integrating

\int{dv}=(-5\omega^2)\int{\sin(\omega t)}dt

v=5\omega^2\dfrac{\cos(\omega t)}{\omega}+C

v=5\omega\cos(\omega t)+C

At t = 0, v = 5ω

5\omega=5\omega\cos(\omega\times0)+C

 C=0

Put the value of C

v=5\omega\cos(\omega t)

\dfrac{dx}{dt}=5\omega\cos(\omega t)

dx=5\omega\cos(\omega t)dt

\int{dx}=5\omega\int{\cos(\omega t)}dt

On integrating again

x=5\omega\dfrac{\sin(\omega t)}{\omega}+C

x=5\sin(\omega t)+C

At t=0, x = 0

0=5\sin(\omega\times 0)+C

C=0

Put the value of C

x=5\sin(\omega t)

Hence, The displacement of this particle at time t is 5\sin(\omega t)

(2) is correct option

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