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If the maximum speed and acceleration of a particle executing SHM is `20 cm//s and 100pi cm//s^2`, find the time period od oscillation.

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To solve the problem, we need to find the time period of oscillation for a particle executing Simple Harmonic Motion (SHM) given its maximum speed and maximum acceleration. ### Step-by-Step Solution: 1. **Identify the Given Values**: - Maximum Speed, \( V_{\text{max}} = 20 \, \text{cm/s} \) - Maximum Acceleration, \( A_{\text{max}} = 100\pi \, \text{cm/s}^2 \) ...
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CENGAGE PHYSICS ENGLISH-LINEAR AND ANGULAR SIMPLE HARMONIC MOTION-Exercise 4.1
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  2. The equation of motion of a particle started at t=0 is given by x=5sin...

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  3. A particle starts SHM from mean position O executing SHM A and B are t...

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  4. If the maximum speed and acceleration of a particle executing SHM is 2...

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  5. A particle is performing SHM of amplitude 'A' and time period 'T'. Fi...

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  6. A particle of mass 2 kg is moving of a straight line under the action ...

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  7. A particle executing simple harmonic motion has amplitude of 1 m and t...

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  8. In the previous question, find maximum velocity and maximum accelerati...

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  9. A partical in SHM ha a period of 4s .It takes time t(1) to start from ...

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  10. A particle is subjected to two simple harmonic motion in the same dire...

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  11. A particle executes SHM of period 1.2 s and amplitude 8cm. Find the ti...

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  12. A cylinder of mass M and radius R is resting on a horizontal platform ...

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  13. The figure shows the displacement-time graph of a particle executing S...

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  14. A body executing SHM has its velocity its 10 cm//sec and7 cm//sec when...

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  15. A body undergoing SHM about the origin has its equation is given by X=...

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  16. The acceleration-displacement (a-X) graph of a particle executing simp...

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  17. A block is kept on a horizontal table. The stable is undergoing simple...

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  18. A linear harmonic oscillator has a total mechanical energy of 200 J ....

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  19. The potential energy of a particle oscillating along x-axis is given a...

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  20. x(1) = 5 sin omega t x(2) = 5 sin (omega t + 53^(@)) x(3) = -...

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