Question:
A body is executing S.H.M. At a displacement \(x\), its potential energy is \(9\;J\) and at a displacement \(y\), its potential energy is \(16\;J\). The potential energy at displacement \((x+y)\) is
A body is executing S.H.M. At a displacement \(x\), its potential energy is \(9\;J\) and at a displacement \(y\), its potential energy is \(16\;J\). The potential energy at displacement \((x+y)\) is
Show Hint
In S.H.M.,
\[
U\propto x^2
\]
Always use the relation
\[
U=\frac12 kx^2
\]
to connect displacement and potential energy.
Updated On: Jun 22, 2026
- \(25\;J\)
- \(5\;J\)
- \(49\;J\)
- \(7\;J\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Write the expression for potential energy in S.H.M.
The potential energy of a particle executing simple harmonic motion is \[ U=\frac12 kx^2 \] where \(k\) is the force constant and \(x\) is the displacement from mean position.
Step 2: Use the given potential energies.
At displacement \(x\), \[ \frac12 kx^2=9 \] \[ kx^2=18 \] At displacement \(y\), \[ \frac12 ky^2=16 \] \[ ky^2=32 \]
Step 3: Find \(x\) and \(y\) in terms of \(k\).
\[ x=\sqrt{\frac{18}{k}} \] \[ y=\sqrt{\frac{32}{k}} \]
Step 4: Find the potential energy at displacement \((x+y)\).
Potential energy at displacement \((x+y)\) is \[ U'=\frac12 k(x+y)^2 \] \[ =\frac12 k(x^2+y^2+2xy) \] Substituting the values, \[ U'=\frac12\left(18+32+2kxy\right) \] Now, \[ xy=\sqrt{\frac{18}{k}}\sqrt{\frac{32}{k}} \] \[ =\frac{\sqrt{576}}{k} \] \[ =\frac{24}{k} \] Thus, \[ 2kxy=2k\left(\frac{24}{k}\right) \] \[ =48 \] Therefore, \[ U'=\frac12(18+32+48) \] \[ =\frac12(98) \] \[ =49\;J \]
Step 5: Final conclusion.
Hence, the potential energy at displacement \((x+y)\) is \[ \boxed{49\;J} \]
The potential energy of a particle executing simple harmonic motion is \[ U=\frac12 kx^2 \] where \(k\) is the force constant and \(x\) is the displacement from mean position.
Step 2: Use the given potential energies.
At displacement \(x\), \[ \frac12 kx^2=9 \] \[ kx^2=18 \] At displacement \(y\), \[ \frac12 ky^2=16 \] \[ ky^2=32 \]
Step 3: Find \(x\) and \(y\) in terms of \(k\).
\[ x=\sqrt{\frac{18}{k}} \] \[ y=\sqrt{\frac{32}{k}} \]
Step 4: Find the potential energy at displacement \((x+y)\).
Potential energy at displacement \((x+y)\) is \[ U'=\frac12 k(x+y)^2 \] \[ =\frac12 k(x^2+y^2+2xy) \] Substituting the values, \[ U'=\frac12\left(18+32+2kxy\right) \] Now, \[ xy=\sqrt{\frac{18}{k}}\sqrt{\frac{32}{k}} \] \[ =\frac{\sqrt{576}}{k} \] \[ =\frac{24}{k} \] Thus, \[ 2kxy=2k\left(\frac{24}{k}\right) \] \[ =48 \] Therefore, \[ U'=\frac12(18+32+48) \] \[ =\frac12(98) \] \[ =49\;J \]
Step 5: Final conclusion.
Hence, the potential energy at displacement \((x+y)\) is \[ \boxed{49\;J} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Energy in simple harmonic motion Questions
- The potential energy of a simple harmonic oscillator of mass $2\, kg$ at its mean position is $5\, J.$ If its total energy is $9\, J$ and amplitude is $1\, cm$. then its time period is
- Position of a \(2\,\text{kg}\) mass moving along the \(x\)-axis is given by \[ x=2\cos[(2\,\text{s}^{-1})t]\ \text{m}. \] Then maximum kinetic energy of the mass in joule is
- If the potential energy at a displacement \(x\) is numerically equal to the square root of potential energy at a displacement \(y\), then (where \(\omega\) is angular velocity):
- A block is executing simple harmonic motion and its position is given by \(x = A \cos(\omega t + \phi)\). At time \(t = 0\), the ratio of the kinetic energy to the potential energy is \(\frac{1}{3}\). The magnitude of the phase angle \(\phi\) is:
- If the amplitude of a lightly damped oscillator decreases by \(1.5\%\), then the mechanical energy of the oscillator lost in each cycle is
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions