14.22 Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

Answers (1)

Let the equation of oscillation be given by x=Asin(\omega t)

Velocity would be given as 

\\v=\frac{dx}{dt}\\ v=A\omega cost(\omega t)

Kinetic energy at an instant is given by

\\K(t)=\frac{1}{2}m(v(t))^{2}\\ K(t)=\frac{1}{2}m(A\omega cos(\omega t))^{2}\\ K(t)=\frac{1}{2}mA^{2}\omega ^{2}cos^{2}\omega t

Time Period  is given by

T=\frac{2\pi }{\omega }

The Average Kinetic Energy would be given as follows

\\K_{av}=\frac{\int _{0}^{T}K(t)dt}{\int _{0}^{T}dt}\\ K_{av}=\frac{1}{T}\int _{0}^{T}K(t)dt\\ K_{av}=\frac{1}{T}\int_{0}^{T}\frac{1}{2}mA^{2}\omega ^{2}cos^{2}\omega t\ dt\\K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\int_{0}^{T}cos^{2}\omega t\ dt\\ K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\int_{0}^{T}\left ( \frac{1+cos2\omega t}{2} \right )dt

\\K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\left [ \frac{t}{2} +\frac{sin2\omega t}{4\omega }\right ]_{0}^{T}\\ K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\left [ \left ( \frac{T}{2}+\frac{sin2\omega T}{4\omega } \right )-\left ( 0+sin(0) \right ) \right ]\\ K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\times \frac{T}{2}\\ K_{av}=\frac{mA^{2}\omega ^{2}}{4}

The potential energy at an instant T is given by 

\\U(t)=\frac{1}{2}kx^{2}\\ U(t)=\frac{1}{2}m\omega ^{2}(Asin(\omega t))^{2}\\ U(t)=\frac{1}{2}m\omega ^{2}A^{2}sin^{2}\omega t

The Average Potential Energy would be given by

\\U_{av}=\frac{\int_{0}^{T}U(t)dt}{\int_{0}^{T}dt}\\ \\U_{av}=\frac{1}{T}\int_{0}^{T}\frac{1}{2}m\omega ^{2}A^{2}sin^{2}\omega t\ dt\\ \\U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\int_{0}^{T}sin^{2}\omega t\ dt\\\\ \\U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\int_{0}^{T}\frac{(1-cos2\omega t)}{2}dt

\\U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\left [ \frac{t}{2} -\frac{sin2\omega t}{4\omega }\right ]_{0}^{T}\\ \\U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\left [ \left ( \frac{T}{2}-\frac{sin2\omega T}{4\omega } \right )-\left ( 0-sin0 \right ) \right ]\\ U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\times \frac{T}{2}\\ U_{av}=\frac{m\omega ^{2}A^{2}}{4}

We can see Kav = Uav

Most Viewed Questions

Related Chapters

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 39999/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions