A particle undergoing simple harmonic motion has a velocity

A particle undergoing simple harmonic motion has a velocity 1, when the displacement is X1 anda velocity %, when the displacement is 1 Find the angular _ frequency and the ampli- tude of the motion in terms of the given quantities. Hint: Since this is SHO, energy is conserved

The Correct Answer and Explanation is:

To solve this, we use the principle of conservation of energy for a particle in Simple Harmonic Motion (SHM). The total mechanical energy remains constant and is the sum of kinetic and potential energy at any point.

Let:

• v1v_1v1​ be the velocity when displacement is x1x_1x1​
• v2v_2v2​ be the velocity when displacement is x2x_2x2​
• ω\omegaω be the angular frequency
• AAA be the amplitude
• mmm be the mass of the particle (it will cancel out, so its value is not needed)

The total energy in SHM is given by:E=12mv2+12mω2×2=12mω2A2E = \frac{1}{2}mv^2 + \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 A^2E=21​mv2+21​mω2×2=21​mω2A2

Step 1: Write energy expressions at two positions

At position x1x_1x1​, the total energy is:12mv12+12mω2×12\frac{1}{2}mv_1^2 + \frac{1}{2}m\omega^2 x_1^221​mv12​+21​mω2×12​

At position x2x_2x2​, the total energy is:12mv22+12mω2×22\frac{1}{2}mv_2^2 + \frac{1}{2}m\omega^2 x_2^221​mv22​+21​mω2×22​

Since total energy is conserved:12mv12+12mω2×12=12mv22+12mω2×22\frac{1}{2}mv_1^2 + \frac{1}{2}m\omega^2 x_1^2 = \frac{1}{2}mv_2^2 + \frac{1}{2}m\omega^2 x_2^221​mv12​+21​mω2×12​=21​mv22​+21​mω2×22​

Divide through by 12m\frac{1}{2}m21​m:v12+ω2×12=v22+ω2x22v_1^2 + \omega^2 x_1^2 = v_2^2 + \omega^2 x_2^2v12​+ω2×12​=v22​+ω2×22​

Step 2: Solve for ω2\omega^2ω2

ω2(x12−x22)=v22−v12\omega^2(x_1^2 – x_2^2) = v_2^2 – v_1^2ω2(x12​−x22​)=v22​−v12​ω2=v22−v12x12−x22\omega^2 = \frac{v_2^2 – v_1^2}{x_1^2 – x_2^2}ω2=x12​−x22​v22​−v12​​

Step 3: Find Amplitude AAA

From total energy:12mω2A2=12mv12+12mω2×12\frac{1}{2}m\omega^2 A^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}m\omega^2 x_1^221​mω2A2=21​mv12​+21​mω2×12​

Divide by 12m\frac{1}{2}m21​m:ω2A2=v12+ω2×12\omega^2 A^2 = v_1^2 + \omega^2 x_1^2ω2A2=v12​+ω2×12​A2=v12ω2+x12A^2 = \frac{v_1^2}{\omega^2} + x_1^2A2=ω2v12​​+x12​

So,A=x12+v12ω2A = \sqrt{x_1^2 + \frac{v_1^2}{\omega^2}}A=x12​+ω2v12​​​

Final Answers:

• Angular frequency:

ω=v22−v12x12−x22\omega = \sqrt{ \frac{v_2^2 – v_1^2}{x_1^2 – x_2^2} }ω=x12​−x22​v22​−v12​​​

• Amplitude:

A=x12+v12ω2A = \sqrt{x_1^2 + \frac{v_1^2}{\omega^2}}A=x12​+ω2v12​​​

This method relies purely on energy conservation and algebraic manipulation, which is valid for any SHM system where velocity and displacement are known at two points.