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Simple Harmonic Motion question

2023 · 1 Feb · Shift 1 · Q70
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  5. /2023 · 1 Feb · Shift 1 · Q70

Simple Harmonic Motion question

2023 · 1 Feb · Shift 1 · Q70

JEE MainPhysicsSimple Harmonic MotionNumerical+4 / −1
The amplitude of a particle executing SHM is 3 cm3 \mathrm{~cm}3 cm. The displacement at which its kinetic energy will be 25%25 \%25% more than the potential energy is: ‾\underline{\hspace{2cm}}​cm\mathrm{cm}cm
Numerical answer
View written solutionFree

Correct answer: 2

  1. Write expressions for energies in SHM

For a particle executing SHM with amplitude AAA, total energy is E=12kA2.E=\frac{1}{2}kA^2.E=21​kA2.

At displacement xxx from mean position:

  • Potential energy, U=12kx2U=\frac{1}{2}kx^2U=21​kx2
  • Kinetic energy, K=E−U=12k(A2−x2).K=E-U=\frac{1}{2}k(A^2-x^2).K=E−U=21​k(A2−x2).
  1. Use the given condition

The kinetic energy is 25%25\%25% more than the potential energy. So, K=1.25 U=54U.K=1.25\,U=\frac{5}{4}U.K=1.25U=45​U.

Substitute the expressions: 12k(A2−x2)=54(12kx2).\frac{1}{2}k(A^2-x^2)=\frac{5}{4}\left(\frac{1}{2}kx^2\right).21​k(A2−x2)=45​(21​kx2).

Cancel 12k\frac{1}{2}k21​k from both sides: A2−x2=54x2.A^2-x^2=\frac{5}{4}x^2.A2−x2=45​x2.

So, A2=(1+54)x2=94x2.A^2=\left(1+\frac{5}{4}\right)x^2=\frac{9}{4}x^2.A2=(1+45​)x2=49​x2.

Hence, x2=49A2.x^2=\frac{4}{9}A^2.x2=94​A2.

Given A=3 cmA=3\text{ cm}A=3 cm, x2=49(3)2=4.x^2=\frac{4}{9}(3)^2=4.x2=94​(3)2=4.

Therefore, x=2 cm.x=2\text{ cm}.x=2 cm.

Since displacement can be on either side of the mean position, the magnitude is 2 cm2\text{ cm}2 cm.

  1. Final answer

2\boxed{2}2​

  1. Comparison with stored answer

Stored correct answer = 222.

My derived answer matches the stored answer.

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