
In damped oscillation mass is 2 Kg and spring constant is $500\,N/m$ and damping coefficient is $1\,Kg/s$. 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
162.6k+ views
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:
$E=\dfrac{1}{2}kx^{2}e^{\frac{-bt}{m}}$
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:
$E=\dfrac{1}{2}kx^{2}e^{\frac{-bt}{m}}$
Putting values from the question in above equation;
$E=\dfrac{1}{2}\times500\times (0.2)^{2}e^{\frac{-1\times4}{2}}$
$E=250\times (0.4)\times e^{-2}$
Further solving, we get;
$E = 100{e^{ - 2}}$
$E = 1.37\,J$
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.
Formula used
Mechanical energy is given by:
$E=\dfrac{1}{2}kx^{2}e^{\frac{-bt}{m}}$
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:
$E=\dfrac{1}{2}kx^{2}e^{\frac{-bt}{m}}$
Putting values from the question in above equation;
$E=\dfrac{1}{2}\times500\times (0.2)^{2}e^{\frac{-1\times4}{2}}$
$E=250\times (0.4)\times e^{-2}$
Further solving, we get;
$E = 100{e^{ - 2}}$
$E = 1.37\,J$
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.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Charging and Discharging of Capacitor

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
