Question

the acceleration of a particle is given by relation is equals to minus 5 omega square Sin Omega Tthe acceleration of a particle is given by relation is equals to minus 5 omega square Sin Omega T at t is equals to zero X is equals to Zero B is equals to 5 Omega the displacement of the particle at time T will be ​

Answer 1

Answer:

s=$5sin\omega t$

Explanation:

We are given that

Acceleration a=$-5\omega^2 sin\omega t$

t=0 and velocity=$5\omega$

$a=\frac{dv}{dt}$

$dv=a\cdo dt$

v=$\int -5w^2sinwtdt$

v=$5w^2(\frac{cos\omega t}{w})$+c

v=$5\omega cos\omega t$ +c

Substitute t=0, then v=$5\omega$

Then we get

$5\omega=5\omega+c$

c=0

v=$5\omega cos\omega t$

$v=\frac{ds}{dt}$

ds=$v\cod dt$

$s=\int(5\omega cos\omega t)dt$

s=$5\omega\frac{sin\omega t}{\omega}+c$

s=$5sin\omega t+c$

When t=0,x=0 then we get

c=0

Hence, the displacement at time t is given by

s=$5sin\omega t$

c