Question

A body of mass $600\text{gm}$ is attached to a spring of spring constant $k=100\text{N/m}$ and it is performing damped oscillations. If damping constant is $0.2$ and driving force is $F=F_0\cos \left(\omega t\right)$ where $F_0=20\text{N}$, find the amplitude of oscillations at resonance.

• A. $4.1\text{m}$
• B. $0.57\text{m}$
• C. $7.75\text{m}$
• D. $0.98\text{m}$

Answer 1

The correct option is C $7.75\text{m}$

We know that, amplitude of oscillation at resonance is given by
$A=\frac{F_0}{2m\gamma {\omega }_0}=\frac{F_0}{b{\omega }_0}$
where $b$ is the damping constant.
From the data given in the question,
$F_0=20\text{N}$
$m=600\text{g}$
$b=0.2\text{rad kg /s}$
Angular frequency of free oscillations
${\omega }_0=\sqrt{\frac{k}{m}}=\sqrt{\frac{100}{0.6}}=12.9\text{rad/sec}$
Now, we have
$A=\frac{20}{12.9\times 0.2}=7.75\text{m}$
Thus, option (c) is the correct answer.