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

2024
Views:

309

Pages:

5
Author:

Jabin F.

# Calculation of angular frequency or time period of oscillation: 1st method → write PE as V(x) find k = d²V/dx² | x=sep ω = √(k/m) T = 2π√(m/k) 2nd method → firstly consider equilibrium → Then displace object by displacement x from equilibrium position and after this, write Fnet = ma → If required, use approximation to show that F = -kx or a = -ω²x 3rd method: In SHM (free oscillation) total energy is conserved. Write total energy at displacement x from equilibrium position and use dE/dt = 0 This gives x'' = -ω²x Q-13 A particle is projected upward with speed 20 m/s. It will return to the initial point after time: (g = 10 m/s²) Solution: The particle will return with same speed but direction is opposite. v² = u² - 2gh ⇒ 0 = 2gh ⇒ h = u²/2g = 400/(2×10) = 20 m Now time for upward motion is same as downward motion. total time = 2T = 2√(2h/g) = 2√(2×20/10) = 2×2 = 4 sec Q-14 A ball is thrown with a speed of 10√2 m/s at 45° to horizontal. If mass of ball is 1 kg, then impulse on the ball during motion is: Solution: u = 10√2 at 45° projectile. ux = 10√2 cos 45° = 10√2 × 1/√2 = 10 uy = 10√2 sin 45° = 10√2 × 1/√2 = 10 Now velocity along x-axis doesn't change but the ball returns with same velocity as thrown upward: ux = 10 uy = -10 v = 10i + 10j; v' = 10i - 10j Impulse = mv' - mv = m(10i - 10j - 10i - 10j) = m(-20j) = -40j = -40 kg m/s downward Q-15 A particle of mass m is thrown with initial speed v₀ in horizontal direction. As the particle moves, a force (drag) F = -kv² acts on it. After what distance its kinetic energy will becomes 1/4th of initial kinetic energy? Solution: According to condition of KE 1/2mv² = 1/4(1/2mv₀²) ⇒ v = v₀/2 Now eom → -kv² = ma ⇒ -kv = m(dv/dt) ⇒ -kdx = mdv ∫₀⁻ᵏ dx = m∫ᵥ₀^(ᵥ₀/₂) dv ⇒ -kx = m(v - v₀) ⇒ x = m(v₀ - v)/k x = m(v₀ - v₀/2)/k = mv₀/2k