Question

If the effective spring constant of a 2500 kg car is 50000 N/m, what is the period of its vibration after hitting a bump?

Answer 1

Given,

• Spring Constant, $\sf{k=50000\ N\,m^{-1}}$

• Mass of car, $\sf{m=2500\ kg}$

The car undergoes an SHM when it hits a bump.

So the time period of the car is,

$\longrightarrow\sf{T=\dfrac{2\pi}{\omega}\quad\quad\dots(1)}$

where $\omega$ is the angular frequency.

But for a simple harmonic motion, the acceleration,

$\longrightarrow\sf{a=-\omega^2x}$

$\longrightarrow\sf{\omega^2=-\dfrac{a}{x}\quad\quad\dots(2)}$

where $\sf{x}$ is the displacement.

But the restoring force,

$\longrightarrow\sf{F=-kx}$

By second law of motion,

$\longrightarrow\sf{ma=-kx}$

$\longrightarrow\sf{-\dfrac{a}{x}=\dfrac{k}{m}}$

Then (2) becomes,

$\longrightarrow\sf{\omega^2=\dfrac{k}{m}}$

$\longrightarrow\sf{\omega=\sqrt{\dfrac{k}{m}}}$

Hence (1) becomes,

$\longrightarrow\sf{T=\dfrac{2\pi}{\sqrt{\dfrac{k}{m}}}}$

$\longrightarrow\sf{T=2\pi\sqrt{\dfrac{m}{k}}}$

Substituting the corresponding values,

$\longrightarrow\sf{T=2\pi\sqrt{\dfrac{2500}{50000}}}$

$\longrightarrow\sf{T=2\pi\sqrt{\dfrac{1}{20}}}$

$\longrightarrow\sf{\underline{\underline{T=\dfrac{\pi}{\sqrt5}\ s}}}$

$\longrightarrow\sf{\underline{\underline{T=1.4\ s}}}$