
A particle starts simple harmonic motion from the mean position. Its amplitude is $a$ and total energy $E$. At an instant its kinetic energy is $\dfrac{3}{4}E$. Its displacement at that instant is:
A) $\dfrac{a}{{\sqrt 2 }}$
B) $\dfrac{a}{2}$
C) $\sqrt 3 \dfrac{a}{2}$
D) Zero
Answer
162.6k+ views
Hint:
To find the values of displacement of the particle doing SHM at an instant asked in the question we can use the concept of kinetic energy of a particle doing SHM.
Formula used:
The particle's kinetic energy with mass $m$, equals:
$K = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})$
Complete step by step solution:
The particle's kinetic energy with mass $m$, equals:
$K = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})$
While potential energy is the power that an object can store due to its position in relation to other things, internal tensions, electric charge, or other circumstances.
$U = \dfrac{1}{2}m{\omega ^2}{x^2}$
As we know that the total energy is sum of the kinetic and potential energy, then we have:
$E = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2}) + \dfrac{1}{2}m{\omega ^2}{x^2} \\$
$E = \dfrac{1}{2}m{\omega ^2}{a^2}\,\,\,\,....(i) \\$
Now, according to the question, the kinetic energy is $\dfrac{3}{4}E$. Then we have:
$\dfrac{3}{4}E = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})\,\,\,\,....(ii)$
Compare the equation $(i)$from $(ii)$, then we obtain:
$1 - \dfrac{{{x^2}}}{{{a^2}}} = \dfrac{3}{4} \\$
$\Rightarrow \dfrac{x}{a} = \sqrt {\dfrac{1}{4}} \\$
$\Rightarrow \dfrac{x}{a} = \dfrac{1}{2} \\$
$\Rightarrow x = \dfrac{a}{2} \\$
Thus, the correct option is:(B) $\dfrac{a}{2}$
Note:
It should be noted that we will have one negative value and one positive value when solving for the displacement. However, we have only thought about the positive root. This is due to the fact that the options only mentioned positive displacements.
To find the values of displacement of the particle doing SHM at an instant asked in the question we can use the concept of kinetic energy of a particle doing SHM.
Formula used:
The particle's kinetic energy with mass $m$, equals:
$K = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})$
Complete step by step solution:
The particle's kinetic energy with mass $m$, equals:
$K = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})$
While potential energy is the power that an object can store due to its position in relation to other things, internal tensions, electric charge, or other circumstances.
$U = \dfrac{1}{2}m{\omega ^2}{x^2}$
As we know that the total energy is sum of the kinetic and potential energy, then we have:
$E = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2}) + \dfrac{1}{2}m{\omega ^2}{x^2} \\$
$E = \dfrac{1}{2}m{\omega ^2}{a^2}\,\,\,\,....(i) \\$
Now, according to the question, the kinetic energy is $\dfrac{3}{4}E$. Then we have:
$\dfrac{3}{4}E = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})\,\,\,\,....(ii)$
Compare the equation $(i)$from $(ii)$, then we obtain:
$1 - \dfrac{{{x^2}}}{{{a^2}}} = \dfrac{3}{4} \\$
$\Rightarrow \dfrac{x}{a} = \sqrt {\dfrac{1}{4}} \\$
$\Rightarrow \dfrac{x}{a} = \dfrac{1}{2} \\$
$\Rightarrow x = \dfrac{a}{2} \\$
Thus, the correct option is:(B) $\dfrac{a}{2}$
Note:
It should be noted that we will have one negative value and one positive value when solving for the displacement. However, we have only thought about the positive root. This is due to the fact that the options only mentioned positive displacements.
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
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

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

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

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

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

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