Question

In damped oscillation mass is 2 Kg and spring constant is $500N/m$ and damping coefficient is $1Kg/s$. If mass is displaced by 20 cm from its mean position and then released what will be value of its mechanical energy after 4 seconds?

A. 2.37 J
B. 1.37 J
C. 10 J
D. 5 J

Answer 1

Hint: First try to find the relation between the mechanical energy, mass, spring constant, damping coefficient, displaced position of mass and the time. After finding the required relation put all the values from the question and finally get the required answer that is the mechanical energy.




Formula used
Mechanical energy is given by:
$E={\displaystyle \frac{1}{2}}k{x}^2{e}^{\frac{-bt}{m}}$
Where, m is the mass of the body.
k is spring constant.
b is the damping coefficient.
t is time.
x is position displaced.



Complete answer:

First start with the given information:
Mass of the body, m = 2 Kg
Spring constant, $k=500N/m$
Damping coefficient, $b=1Kg/s$
Time, $t=4\sec$
Position displaced, $x=20cm$
We know that the mechanical energy in case of damping oscillation:
$E={\displaystyle \frac{1}{2}}k{x}^2{e}^{\frac{-bt}{m}}$
Putting values from the question in above equation;
$E={\displaystyle \frac{1}{2}}\times 500\times (0.2{)}^2{e}^{\frac{-1\times 4}{2}}$
$E=250\times (0.4)\times e^{-2}$
Further solving, we get;
$E=100e^{-2}$
$E=1.37J$
Hence, the correct answer is Option B.




Note:Here in order to find the mechanical energy in case of damped oscillation all the values were already given in the question so we just have to put all the values and get the required answer, if any of the value is missing in any other case then the answer will differ in that case.