Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A linear harmonic oscillator of force constant <img alt="\mathrm{2 \times 10^6 \mathrm{~N} / \mathrm{m}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B2%20%5Ctimes%2010%5E6%20%5

A linear harmonic oscillator of force constant \mathrm{2 \times 10^6 \mathrm{~N} / \mathrm{m}} and amplitude \mathrm{0.01 \mathrm{~m}} has a total mechanical energy \mathrm{ 160 \mathrm{~J}}. Find the maximum and minimum value of potential energy \mathrm{ (PE) and K E.}
 

Option: 1

\mathrm{k_{\text {max }}=100 \mathrm{~J}, P E_{\text {max }}=160 \mathrm{~J}}



 


Option: 2

\mathrm{k_{\text {max }}=200 \mathrm{~J}, P E_{\text {max }}=170 \mathrm{~J}}


Option: 3

\mathrm{k_{\text {max }}=120 \mathrm{~J}, P E_{\text {max }}=160 \mathrm{~J}}


Option: 4

\mathrm{k_{\text {max }}=160 \mathrm{~J}, k_{\text {max }}=100 \mathrm{~J}}


Answers (1)

best_answer

Use, \mathrm{k_{\text {max }}=\frac{1}{2} k A^2=\frac{1}{2} \times\left(2 \times 10^6\right) \times(0.01)^2}

                                     \mathrm{ =100 \mathrm{~J} }

using,\mathrm{ k_{\text {max }}+U_{\min }=160 }

         \mathrm{ 100+U_{\min }=160 \quad\quad \therefore U_{\min }=60 \mathrm{~J} }

Maximum \mathrm{ P E= } total mechanical energy.

                       \mathrm{ =160\: J }

Hence option 1 is correct.

Posted by

qnaprep

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Mains 2026 April 5 Shift 1: Students find Maths lengthy, time-consuming; Physics moderate, Chemistry easy
anam-khan

JEE Mains 2026 April 5 LIVE: Day 3 Shift 1 BTech, BE exam; Maths 'challenging'; physics, chemistry ‘balanced'
anam-khan

JEE Mains 2026 April 4 Shift 2: Maths remains ‘moderate, lengthy’ in difficulty; physics, chemistry ‘balanced'

Latest Question