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

2024
Views:

216

Pages:

5
Author:

Jabin F.

# Spring Mass System (Heavy Spring) Here spring also has mass. Since spring has mass, so moving turns of spring also have kinetic energy. Every turn have different speed, so KE of the spring must be calculated by summing all the KE. [Diagram of a spring-mass system with a heavy spring] Let a small element dy, y distance apart from fixed point having mass dm = (m/L)dy. KE of small element = (1/2)dm * v² = (1/2)(m/L)dy * v² Assume linear variation of speed of turns of spring: v = (y/L) * v₀ KE of small element = (1/2)(m/L)dy * ((y/L) * v₀)² = (1/2)(m/L³) * v₀² * y³ dy Total KE of spring = ∫₀ˡ (1/2)(m/L³) * v₀² * y³ dy = (1/8)mv₀² KE of the block = (1/2)mv₀² PE of the spring = (1/2)kx² Total Energy = (1/2)mv₀² + (1/8)mv₀² + (1/2)kx² Apply conservation of energy: (1/2)mv₀² + (1/8)mv₀² + (1/2)kx² = 0 (5/8)mv₀² + (1/2)kx² = 0 (5/4)mv₀² + kx² = 0 x² = -(5/4)(mv₀²/k) x = ±√(-5/4)(mv₀²/k) Now (T^{2}=\frac{4\pi^{2}}{k}(m+\frac{m_{s}}{3})) [T^{2}=\frac{4\pi^{2}}{k}m+\frac{4\pi^{2}m_{s}}{3~k}] as (y=mx+c) where (m=tan~\theta=\frac{4\pi^{2}}{k}] Vary m and note T to get a graph between m and T². [Diagram of a graph between m and T²] The spring constant k, of a spring of mass ms, is determined experimentally by loading the spring with mass m and recording time period T, for a single oscillation. If the experiment is carried out for different masses, then the graph that correctly represents the result is: [Diagram of four graphs representing the relationship between m and T²] Solution: As described in the previous article. P-54 Two particles of equal masses and opposite charges are fixed at a distance d apart. Due to electrostatic force, they move towards each other. When the separation between them reduces to d/4, their speeds become u. At what separation will the speed of each particle be 2u? Solution: Since all forces are conservative: So apply conservation of energy: (KE + PE)initial = (KE + PE)final (1/2)m(0)² + (1/2)m(0)² - (q²/4πε₀d) = (1/2)mu² + (1/2)mu² - (q²/4πε₀(d/4)) (1/2)mu² = (q²/4πε₀d) * (1 - 1/4) (1/2)mu² = (3/4)(q²/4πε₀d) mu² = (3/2)(q²/4πε₀d) Again apply conservation of energy: (KE + PE)initial = (KE + PE)final (1/2)m(0)² + (1/2)m(0)² - (q²/4πε₀d) = (1/2)m(2u)² + (1/2)m(2u)² - (q²/4πε₀x) (1/2)m(4u²) = (q²/4πε₀d) * (1 - 1/x) (2mu²) = (q²/4πε₀d) * (1 - 1/x) From the previous equation, we know: mu² = (3/2)(q²/4πε₀d) Substitute this value: (2 * (3/2)(q²/4πε₀d)) = (q²/4πε₀d) * (1 - 1/x) (3/2) = 1 - 1/x 1/x = 1 - (3/2) 1/x = -1/2 x = -2