Question:
A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air jet when it is at 5 cm from its mean position. The new amplitude of vibration is \(\sqrt x\) cm. The value of x is ___.
A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air jet when it is at 5 cm from its mean position. The new amplitude of vibration is \(\sqrt x\) cm. The value of x is ___.
Updated On: Mar 19, 2025
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Correct Answer: 700
Solution and Explanation
The correct option is: 700
\(v=ω\sqrt{A^2-y^2}\)
⇒ 9 × 75 = (A′)2 – 25
\(⇒A'=\sqrt{28×25}\,cm\)
⇒ x = 700
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Concepts Used:
Energy In Simple Harmonic Motion
We can note there involves a continuous interchange of potential and kinetic energy in a simple harmonic motion. The system that performs simple harmonic motion is called the harmonic oscillator.
Case 1: When the potential energy is zero, and the kinetic energy is a maximum at the equilibrium point where maximum displacement takes place.
Case 2: When the potential energy is maximum, and the kinetic energy is zero, at a maximum displacement point from the equilibrium point.
Case 3: The motion of the oscillating body has different values of potential and kinetic energy at other points.