The amplitude of a particle executing SHM is 3 cm. The displacement at which its kinetic energy will be 25% more than the potential energy is ______ cm. [2023]

Question

The amplitude of a particle executing SHM is 3 cm. The displacement at which its kinetic energy will be 25% more than the potential energy is ______ cm. [2023]

Smart Achievers · Accepted Answer

(2)

$K E = P E + \frac{P E}{4}$

$K E = \frac{5}{4} P E$

$\frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{5}{4} \times \frac{1}{2} m ω^{2} x^{2}$

$[v = ω \sqrt{A^{2} - x^{2}}]$

$A^{2} - x^{2} = \frac{5}{4} x^{2}$

$\frac{9 x^{2}}{4} = A^{2} \Rightarrow x = \frac{2}{3} A$

$∴$ $x = \frac{2}{3} \times 3 cm \Rightarrow x = 2 cm$

Solution

Q.
The amplitude of a particle executing SHM is 3 cm. The displacement at which its kinetic energy will be 25% more than the potential energy is ______ cm. [2023]

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Solution

Q. The amplitude of a particle executing SHM is 3 cm. The displacement at which its kinetic energy will be 25% more than the potential energy is ______ cm. [2023]

Ans. (2) KE=PE+PE4 KE=54 PE 12mω2(A2-x2)=54×12mω2x2 [v=ωA2-x2 ] A2-x2=54x2 9x24=A2⇒x=23A ∴ x=23×3 cm⇒x=2 cm

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Q.
The amplitude of a particle executing SHM is 3 cm. The displacement at which its kinetic energy will be 25% more than the potential energy is ______ cm. [2023]