Question

A block of mass $0.1kg$ is connected to an elastic spring of spring constant $640N{m}^{-1}$ 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. $5s$
• B. $7s$
• C. $3.5s$
• D. $2s$

Answer 1

Answer: (B)

$7s$

Displacement of this system is given as

$r=A{e}^{-\lambda t}6s(\omega t-\delta )$

$\lambda =\frac{R}{2m}$

$R={10}^{-2}kg/s$

$\lambda =\frac{{10}^{-2}}{2\times 0.1}=0.05s^{-1}$

Energy$=\frac{1}{2}k{x}^2$ $k$ is spring constant

$E=\frac{1}{2}k{A}^2{e}^{-2\lambda t}{\cos }^2(\omega t-\delta )$

Suppose $\delta =0E_0=\frac{K{A}^2}{2}$

$E_{\frac{1}{2}}=\frac{K{A}^2}{4}=\frac{K{A}^2}{2}{e}^{-2\lambda t}{\cos }^2\left(\omega t\right)$

$0.5=e^{-0.1t}{\cos }^2\left(\omega t\right)$---------(1)

$\omega =\sqrt{\frac{m}{k}}=\sqrt{\frac{0.1}{640}}=0.0125$

By putting the value of $\omega$ in (1) we get,

$t\simeq 7seconds$