Question:
Assertion (A): In S.H.M., kinetic and potential energies become equal when the distance is
\[
\frac{1}{\sqrt{2}}
\]
times the amplitude.
Reason (R): The potential energy of a particle executing S.H.M. is periodic with time and is maximum at the extreme displacement.
Assertion (A): In S.H.M., kinetic and potential energies become equal when the distance is
\[
\frac{1}{\sqrt{2}}
\]
times the amplitude.
Reason (R): The potential energy of a particle executing S.H.M. is periodic with time and is maximum at the extreme displacement.
Show Hint
For S.H.M.,
\[
U=\frac{1}{2}kx^2,
\qquad
K=\frac{1}{2}k(A^2-x^2).
\]
Whenever
\[
K=U,
\]
the displacement is
\[
x=\frac{A}{\sqrt2}.
\]
This is a very common result in S.H.M. problems.
Updated On: Jun 26, 2026
- (A) and (R) are true. (R) is the correct explanation of (A)
- (A) and (R) are true. (R) is not the correct explanation of (A)
- (A) is true, but (R) is false
- (A) is false, but (R) is true
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Verify Assertion (A).
For a particle executing S.H.M., the total energy is \[ E=\frac{1}{2}kA^2, \] where \(A\) is the amplitude.
The potential energy at displacement \(x\) is \[ U=\frac{1}{2}kx^2. \] The kinetic energy is \[ K=E-U = \frac{1}{2}kA^2-\frac{1}{2}kx^2. \] When kinetic energy equals potential energy, \[ K=U. \] Hence, \[ \frac{1}{2}kA^2-\frac{1}{2}kx^2 = \frac{1}{2}kx^2. \] \[ A^2=2x^2. \] \[ x=\frac{A}{\sqrt2}. \] Therefore, \[ \boxed{x=\frac{A}{\sqrt2}}. \] Hence Assertion (A) is true.
Step 2: Verify Reason (R).
The potential energy in S.H.M. is \[ U=\frac{1}{2}kx^2. \] Since displacement varies periodically with time, potential energy also varies periodically with time.
At the extreme positions, \[ x=\pm A. \] Therefore, \[ U_{\max} = \frac{1}{2}kA^2. \] Thus, the potential energy is maximum at the extreme displacement.
Hence Reason (R) is also true.
Step 3: Check whether (R) explains (A).
Assertion (A) is obtained from the condition \[ K=U \] and the energy relations \[ K=\frac{1}{2}k(A^2-x^2), \qquad U=\frac{1}{2}kx^2. \] Reason (R) only states that potential energy is periodic and maximum at the extreme positions.
Although true, it does not explain why \[ K=U \] occurs at \[ x=\frac{A}{\sqrt2}. \] Therefore, Reason (R) is not the correct explanation of Assertion (A).
Step 4: Final conclusion.
Thus, \[ \boxed{\text{(A) and (R) are true, but (R) is not the correct explanation of (A).}} \] Hence, the correct option is \[ \boxed{(2)} \]
For a particle executing S.H.M., the total energy is \[ E=\frac{1}{2}kA^2, \] where \(A\) is the amplitude.
The potential energy at displacement \(x\) is \[ U=\frac{1}{2}kx^2. \] The kinetic energy is \[ K=E-U = \frac{1}{2}kA^2-\frac{1}{2}kx^2. \] When kinetic energy equals potential energy, \[ K=U. \] Hence, \[ \frac{1}{2}kA^2-\frac{1}{2}kx^2 = \frac{1}{2}kx^2. \] \[ A^2=2x^2. \] \[ x=\frac{A}{\sqrt2}. \] Therefore, \[ \boxed{x=\frac{A}{\sqrt2}}. \] Hence Assertion (A) is true.
Step 2: Verify Reason (R).
The potential energy in S.H.M. is \[ U=\frac{1}{2}kx^2. \] Since displacement varies periodically with time, potential energy also varies periodically with time.
At the extreme positions, \[ x=\pm A. \] Therefore, \[ U_{\max} = \frac{1}{2}kA^2. \] Thus, the potential energy is maximum at the extreme displacement.
Hence Reason (R) is also true.
Step 3: Check whether (R) explains (A).
Assertion (A) is obtained from the condition \[ K=U \] and the energy relations \[ K=\frac{1}{2}k(A^2-x^2), \qquad U=\frac{1}{2}kx^2. \] Reason (R) only states that potential energy is periodic and maximum at the extreme positions.
Although true, it does not explain why \[ K=U \] occurs at \[ x=\frac{A}{\sqrt2}. \] Therefore, Reason (R) is not the correct explanation of Assertion (A).
Step 4: Final conclusion.
Thus, \[ \boxed{\text{(A) and (R) are true, but (R) is not the correct explanation of (A).}} \] Hence, the correct option is \[ \boxed{(2)} \]
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 Simple Harmonic Motion Questions
- A body is executing simple harmonic motion. At a displacement \( x \), its potential energy is \( E_1 \), and at a displacement \( y \), its potential energy is \( E_2 \). The potential energy \( E \) at a displacement \( (x + y) \) is:
- A block A of mass 1.6 kg is executing SHM with amplitude of 10 cm. A block B of mass 0.9 kg is kept on A. When A passes through its equilibrium, the amplitude of oscillation of the system of block A and B is
- A particle is executing simple harmonic motion. If the minimum time taken by the particle to move from extreme position to half of the amplitude is \( t_1 \), and the minimum time taken by the particle to move from mean position to half of the amplitude is \( t_2 \), then
- When a body of mass 8 kg is attached to a spring balance, the reading of the balance is 20 cm. Instead of 8 kg, if another body of mass M is suspended from the spring balance and is made to oscillate vertically, the time period of oscillation is \( \frac{\pi}{5} \) s, then the value of M is (Acceleration due to gravity = 10 ms\(^{-2} \))
- Under the action of a force \( F = -75y \) where F is in Newton and y is in meters, an object of mass 3 kg executes simple harmonic motion. If the velocity of the object at the mean position is 2.5 ms\(^{-1} \), the maximum acceleration of the object is
View More Questions
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