Question 140: Oscillations

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 :

Accepted Answer

To determine the time taken for the mechanical energy of the damped oscillator to drop to half its initial value, we'll use the principles of damped harmonic motion.
Given:
Mass of the block ($ m $)
: 0.1Ã¢ÂÂ¯kg
Spring constant ($ k $)
: 640Ã¢ÂÂ¯N/m
Damping constant ($ b $)
: $ 10^{-2} $Ã¢ÂÂ¯kg/s
Understanding Damped Harmonic Motion:
In damped harmonic motion, the amplitude of oscillation decreases exponentially over time due to the damping force. The mechanical energy ($ E $) of the oscillator is proportional to the square of its amplitude ($ A $):
$ E(t) \propto A(t)^2 $
The amplitude as a function of time is given by:
$ A(t) = A_0 \, e^{- \frac{b}{2m} t} $
Where:
$ A_0 $ is the initial amplitude.
$ b $ is the damping constant.
$ m $ is the mass.
$ t $ is the time.
Therefore, the mechanical energy as a function of time is:
$ E(t) = E_0 \, e^{- \frac{b}{m} t} $
Where $ E_0 $ is the initial mechanical energy.
Calculating the Time When Energy Drops to Half:
We need to find the time $ t $ when $ E(t) = \frac{1}{2} E_0 $:
$ \frac{1}{2} E_0 = E_0 \, e^{- \frac{b}{m} t} $
Simplify:
$ \frac{1}{2} = e^{- \frac{b}{m} t} $
Take the natural logarithm of both sides:
$ \ln\left(\frac{1}{2}ight) = - \frac{b}{m} t $
Simplify $ \ln\left(\frac{1}{2}ight) = -\ln(2) $:
$ - \ln(2) = - \frac{b}{m} t $
Cancel negatives:
$ \ln(2) = \frac{b}{m} t $
Solve for $ t $:
$ t = \frac{m}{b} \ln(2) $
Plugging in the Given Values:
$ t = \frac{0.1\, 	ext{kg}}{10^{-2}\, 	ext{kg/s}} \ln(2) $
Calculate $ \ln(2) $:
$ \ln(2) \approx 0.6931 $
Now compute $ t $:
$ t = \frac{0.1}{0.01} 	imes 0.6931 = 10 	imes 0.6931 = 6.931\, 	ext{s} $
Conclusion:
The time taken for the mechanical energy to drop to half its initial value is approximately
6.93 seconds
, which is closest to
7 seconds
among the given options.

Answer

2 s

Answer

3.5 s

Answer

5 s

Answer

7 s

Oscillations

Choose the correct answer: