Question:
The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) starting from mean position to extreme position (A) is given by:
The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) starting from mean position to extreme position (A) is given by:
Show Hint
In simple harmonic motion, the kinetic energy is proportional to the square of the displacement, hence the graph of kinetic energy vs displacement is parabolic.
Updated On: Mar 22, 2025
- Parabola
- Straight line
- Kinetic energy and displacement are inversely proportional
- Sine curve
Hide Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
The kinetic energy \( KE \) of a particle executing simple harmonic motion is related to its displacement \( x \) by the following equation:
\[
KE = E - P - E
\]
Where \( E \) is the total mechanical energy and \( P \) is the potential energy. Since \( KE = \frac{1}{2} k x^2 \), the graph of \( KE \) vs \( x \) is a parabola.
Thus, the correct answer is a parabola.
Thus, the correct answer is a parabola.
Was this answer helpful?
0
0
Top Questions on Waves
- Displacement of a wave is expressed as $$ x(t) = 5 \cos \left( 628t + \frac{\pi}{2} \right) \, \text{m}. $$ The wavelength of the wave when its velocity is 300 m/s is:
- The equation of a wave travelling on a string is $ y = \sin[20\pi x + 10\pi t] $, where x and t are distance and time in SI units. The minimum distance between two points having the same oscillating speed is :
- The amplitude and phase of the wave when two travelling waves given as $ y_1(x, t) = 4 \sin(\omega t - kx) $ and $ y_2(x, t) = 2 \sin\left(\omega t - kx + \frac{2\pi}{3}\right) $ are superimposed.
- A light wave is propagating with plane wave fronts of the type \( x + y + z = \text{constant} \). The angle made by the direction of wave propagation with the \( x \)-axis is:
Two plane polarized light waves combine at a certain point, whose "E" components are: \[ E_1 = E_0 \sin \omega t, \quad E_2 = E_0 \sin \left( \omega t + \frac{\pi}{3} \right) \] Find the amplitude of the resultant wave.
View More Questions
Questions Asked in JEE Main exam
- Two equal sides of an isosceles triangle are along \( -x + 2y = 4 \) and \( x + y = 4 \). If \( m \) is the slope of its third side, then the sum of all possible distinct values of \( m \) is:
- JEE Main - 2025
- 3D Geometry
- Match List - I with List - II. List - I (Partial Derivatives) \(and\) List - II (Thermodynamic Quantity)
In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
- Thermodynamics
- Let $ \vec{a} $ and $ \vec{b} $ be the vectors of the same magnitude such that $ \frac{| \vec{a} + \vec{b} | + | \vec{a} - \vec{b} |}{| \vec{a} + \vec{b} | - | \vec{a} - \vec{b} |} = \sqrt{2} + 1. \quad \text{Then } \frac{| \vec{a} + \vec{b} |^2}{| \vec{a} |^2} \text{ is:} $
- JEE Main - 2025
- Vector Algebra
- Let the distance between two parallel lines be 5 units and a point \( P \) lies between the lines at a unit distance from one of them. An equilateral triangle \( POR \) is formed such that \( Q \) lies on one of the parallel lines, while \( R \) lies on the other. Then \( (QR)^2 \) is equal to _____________.
- JEE Main - 2025
- Coordinate Geometry
- The sum of all local minimum values of the function \( f(x) \) as defined below is:
\[ f(x) = \left\{ \begin{array}{ll} 1 - 2x & \text{if } x < -1 \\ \frac{1}{3}(7 + 2|x|) & \text{if } -1 \leq x \leq 2 \\ \frac{11}{18} (x-4)(x-5) & \text{if } x > 2 \end{array} \right. \]
- JEE Main - 2025
- Functions
View More Questions