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A particle executing simple harmonic mot...

A particle executing simple harmonic motion has amplitude of `1 m` and time period `2 s`. At `t = 0`, net force on the particle is zero. Find the equation of displacement of the particle.

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To find the equation of displacement for a particle executing simple harmonic motion (SHM), we can follow these steps: ### Step-by-Step Solution: 1. **Identify Given Values:** - Amplitude (A) = 1 m - Time period (T) = 2 s ...
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CENGAGE PHYSICS ENGLISH-LINEAR AND ANGULAR SIMPLE HARMONIC MOTION-Exercise 4.1
  1. Suppose a tunnel is dug along a diameter of the earth. A particle is d...

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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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