Question

A block of mass $0.1\, kg$ is connected to an elastic spring of spring constant $640 \, Nm^{-1}$ and oscillates in a damping medium of damping constant $10^{-2} \, kg \, s^{-1}$. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to :

• A. 2 s
• B. 3.5 s
• C. 5 s
• D. 7 s

Answer 1

Answer: (D)

Displacement of this system is given as
$
\begin{array}{l}
r = Ae ^{-\lambda t } 6 s (\omega t -\delta) \\
\lambda=\frac{ R }{2 m } \\
R =10^{-2} kg / s \\
\lambda=\frac{10^{-2}}{2 \times 0.1}=0.05 s ^{-1}
\end{array}
$
Energy $=\frac{1}{2} kx ^{2} k$ is spring constant
$
E =\frac{1}{2} kA ^{2} e ^{-2 \lambda t } \cos ^{2}(\omega t -\delta)
$
Suppose $\delta=0$
$
\begin{array}{l}
E _{0}=\frac{ KA ^{2}}{2} \\
E _{\frac{1}{2}}=\frac{ KA ^{2}}{4}=\frac{ KA ^{2}}{2} e ^{-2 \lambda t } \cos ^{2}(\omega t ) \\
0.5= e ^{-0.1 t } \cos ^{2}(\omega t )- \\
\omega=\sqrt{\frac{ m }{ k }}=\sqrt{\frac{0.1}{640}}=0.0125
\end{array}
$
By putting the value of $\omega$ in (1) we get, $t \simeq 7$ seconds