Question

A 280-g mass is mounted on a spring of constant k = 3.3 N/m. The damping constant for this damped system is b = 8.4 x 10^-3 kg/s. How many oscillations will the system undergo during the time the amplitude decays to 1/e of its original value?

Answer 1

Answer:

The number of oscillation is 36.

Explanation:

Given that

Mass = 280 g

Spring constant = 3.3 N/m

Damping constant $b=8.4\times10^{-3}\ Kg/s$

We need to check the system is under-damped, critical damped and over damped by comparing b with $2m\omega_{0}$

$2m\omega_{0}=2m\sqrt{\dfrac{k}{m}}$

$2\sqrt{km}=2\times\sqrt{3.3\times280\times10^{-3}}=1.92kg/s$

Here, b<<$2m\omega_{0}$

So, the motion is under-damped and will oscillate

$\omega=\sqrt{\omega_{0}^2-\dfrac{b^2}{4m^2}}$

The number of oscillation before the amplitude decays to $\dfrac{1}{e}$ of its original value

$A exp(\dfrac{-b}{2m}t)=A exp(-1)$

$\dfrac{b}{2m}t=1$

$t = \dfrac{2m}{b}$

$t=\dfrac{2\times280\times10^{-3}}{8.4\times10^{-3}}$

$t=66.67\ s$

We need to calculate the time period of one oscillation

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\times3.14}{\sqrt{\omega_{0}^2-\dfrac{b^2}{4m^2}}}$

$T=\dfrac{2\times3.14}{\sqrt{\dfrac{k}{m}-\dfrac{b^2}{4m^2}}}$

$T=\dfrac{2\times3.14}{\sqrt{\dfrac{3.3}{280\times10^{-3}}-\dfrac{(8.4\times10^{-3})^2}{4\times(280\times10^{-3})^2}}}$

$T=1.83\ sec$

The number of oscillation is

$n=\dfrac{t}{T}$

$n=\dfrac{66.67}{1.83}$

$n=36$

Hence, The number of oscillation is 36.