Acceleration of a particle varies with distance as $-kx$. The particle is initially given a velocity $v_0$. Find time take to reach initial point

Question

$newcommand{d}{mathrm{d}}$I am having a problem here because acceleration varies with distance, and I am not able to relate it with time
begin{align}
a &=-frac{kx}{m} \
vfrac{d v}{d x} &= -frac{kx}{m}
end{align}
For $x_{max}$
begin{align}
0-v_{0}^2 &= -frac{kx^2}{m} \
x &= sqrt{frac{mv_0}{k}}
end{align}
But I still can’t find time

Tavish · Answer

$newcommand{d}{mathrm{d}}$We have,
$$
v ,d v =-Cx ,d x \
int_{v_0}^{v(x)} v ,d v = -Cint_0^x x,d x \
frac{v(x)^2-v_0^2}{2}=-Cx \
v(x)=sqrt{v_0^2-2Cx},
$$
where $C=frac km$. Using $v=frac{d x}{d t}$, $$frac{d x}{sqrt{v_0^2-2Cx}} = d t$$ Now just integrate both sides to get an expression for $x(t)$, after which you need to solve $x(t)=0$.

Acceleration of a particle varies with distance as $-kx$. The particle is initially given a velocity $v_0$. Find time take to reach initial point

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