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(A) : Projection of a uniform circular m...

(A) : Projection of a uniform circular motion on the diameter of the circle is simple harmonic.
(B) : The energy of a body executing SHM is directly proportional to square of its amplitude.

A

Both A and B are correct

B

Both A and B are wrong

C

A is correct but B is wrong

D

A is wrong but B is correct

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AI Generated Solution

The correct Answer is:
To solve the question, we need to analyze both statements (A) and (B) regarding simple harmonic motion (SHM) and the energy of a body executing SHM. ### Step-by-Step Solution: **Step 1: Analyze Statement (A)** Statement (A) claims that the projection of a uniform circular motion on the diameter of the circle is simple harmonic motion. - **Explanation**: - Consider a point moving in a circle with uniform speed. If we project this point onto a diameter of the circle, the projection will oscillate back and forth along the diameter. - As the point moves around the circle, its projection on the diameter will move from one extreme (maximum displacement) to the other extreme and back, creating a to-and-fro motion. - This motion is periodic and can be described by a sine or cosine function, which is characteristic of SHM. **Conclusion for (A)**: Statement (A) is correct. --- **Step 2: Analyze Statement (B)** Statement (B) states that the energy of a body executing SHM is directly proportional to the square of its amplitude. - **Explanation**: - In SHM, the total mechanical energy (E) is the sum of kinetic energy (KE) and potential energy (PE). - The potential energy in SHM can be expressed as \( PE = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the mean position. - The maximum displacement (amplitude) is denoted by \( A \). At maximum displacement, \( x = A \), thus the potential energy becomes \( PE = \frac{1}{2} k A^2 \). - The kinetic energy varies, but the total energy remains constant and can be expressed as \( E = \frac{1}{2} k A^2 \). - Therefore, the total energy is directly proportional to the square of the amplitude \( A^2 \). **Conclusion for (B)**: Statement (B) is correct. --- ### Final Conclusion: Both statements (A) and (B) are correct. ---
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NARAYNA-OSCILLATIONS-EXERCISE - II (C.W)
  1. The displacement time graph of a particle executing SHM is as shown in...

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  2. Two particles undergo SHM along the same line with the same time perio...

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  3. (A) : Projection of a uniform circular motion on the diameter of the c...

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  4. A particle is executing SHM. Then the graph of acceleration as a funct...

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  5. A body of mass 0.01 kg executes simple harmonic motion (SHM) about x=0...

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  6. The oscillation of a body on a smooth horizontal surface is represente...

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  7. The x-t graph of a particle undergoing simple harmonic motion is shown...

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  8. The displacement - time graph of a particle executing SHM is as shown ...

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  9. The acceleration of the particle at t = 3 s in the above figure is

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  10. The minimum time the particle takes to travel from y = + "1 m to y" = ...

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  11. Match the following {:("List - I","List - II"),("(a) acceleration","...

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  12. For a particle executing SHM along a straight line (displacement is me...

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  13. {:("List - I","List - II"),("(A) x-t graph of simple harmonic oscillat...

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  14. The mass and diameter of a planet are twice those of earth. What will ...

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  15. The length of a second's pendulum on the surface of the moon, where g ...

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  16. The matallic bob of a simple pendulum has the relative density rho. Th...

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  17. A pendulum clock is taken 1km inside the earth from mean sea level. Th...

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  18. A seconds pendulum is suspended from rof of a vehicle that is moving a...

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  19. For a simple pendulum, the graph between T^(2) and L (where T is the t...

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  20. In case of a simple pendulum, time period versus length is depicted by

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