University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

345

Pages:

5
Author:

Jabin F.

Self Inductance (L): L = μ₀ / 2π * ln(b/a) Self Inductance per unit length (for long coaxial cable): l = μ₀ / 2π * ln(b/a) (H/m) Diagram: Coaxial Cable (Transmission Line): A diagram of a coaxial cable is shown. Graph: A graph showing the variation of velocity (v) with displacement (x). Four options (a, b, c, d) for the graph representing the variation of acceleration (a) with displacement. Solution: Note: We can't tell about the graph directly. Always find out the equation of the graph first, then plot it. Given graph is a straight line graph with y-intercept v₀ and let the slope be m, then the equation of the given graph is: y = mx + C General equation of a straight line: y = mx + C Equation of a straight line when x and y intercepts are given: x/a + y/b = 1 Equation of a straight line when two points are given: (y - y₁) = (y₂ - y₁)/(x₂ - x₁) * (x - x₁) In the given plot, x and y intercepts are given, so we use the intercept form of the equation: x/x₀ + y/v₀ = 1 To find acceleration from velocity and displacement: a = dv/dt = dv/dx * dx/dt = v₀ * (-1/x₀) * (dx/dt) = -v₀/x₀ * v = -v₀²/x₀ + v₀ This equation is similar to y = mx + C, where the intercept is negative and the slope is positive. This condition only satisfied by option (A) Ans: Diagram: A diagram showing a parabolic path y = x² and the acceleration vector a. Problem: A particle moves along a parabolic path y = x² with constant speed 2 m/s. What is the acceleration of the particle at (0, 0)? Solution: v = 2 = √(ẋ² + ẏ²) ẋ² + ẏ² = 4 (Equation 1) Differentiating equation 1 with respect to time: 2ẋẍ + 2ẏÿ = 0 ẋẍ + ẏÿ = 0 (Equation 2) Given path: y = x² Differentiating y = x²: ẏ = 2ẋ Differentiating ẏ = 2ẋ: ÿ = 2ẍ² + 2ẋẍ At (0, 0): ẏ = 0 ÿ = 2ẍ² Putting ÿ = 0 in equation 2: ẋẍ = 0 ẋ = 0 or ẍ = 0 Putting ẋ = 0 in ÿ = 2ẍ²: ÿ = 0 Putting ẋ = ±2 and ẏ = 0 in equation 1: ÿ = 2(±2)² = 8 Putting ẋ = ±2, ẏ = 0, and ÿ = 8 in equation 2: ẍ = 0 Net acceleration: a⃗ = ax̂i + aŷj = 0î + 8ĵ = 8ĵ |a⃗| = 8