Energy of SHM

# Energy of a SHO (any oscillation undergoing SHM) E = KE + PE = ½mv² + ½kx² = ½mω²(A² - x²) + ½mω²x² KE PE PE KE A A E = ½mω²A² as ω = √(k/m) => k = mω² Q-8 A particle of unit mass is released at unstable equilibrium position of potential V(x) = x - x³/3. Speed of the particle when it reaches stable equilibrium position is: Solution: V(x) = x - x³/3 V'(x) = 1 - x² = 0 => x = ±1 V''(x) = -2x → at x = 1, V''(x) = -2 < 0 max. UIEP xx = -1 U''(x) = 2 tue min. SEP V(1) = 1 - 1/3 = 2/3 V(-1) = -1 + 1/3 = -2/3 Now apply conservation of energy. (KE + PE)unstable equilibrium position = (KE + PE)stable equilibrium position 0 + 2/3 = ½v² - 2/3 v² = 2(2/3 + 2/3) = 8/3 => v = √(8/3) Q-9 Consider rain drops falling through with terminal velocity vT. Two raindrops of same size coalesce into one while falling, the terminal velocity of this new droplet is: Solution: Eom: mg - 6πηrvT = ma = 0 for terminal velocity, a = 0 mg - 6πηrvT = 0 => vT = mg / 6πηr Now mass of two new raindrop add up to form new droplet: volume of two rain drop add up to form new droplet: 4/3πR³ = 2 * 4/3πr³ => R = 2^(1/3)r vT_new = Mg / 6πηR = 9m / 6πη(2^(1/3)r) = (1/2)^(1/3) * 9vT = (1/2)^(1/3) * 4vT Q-10 A particle is thrown upward with speed √(GM/R), where M and R are mass and radius of earth resp. At what distance from centre of earth its kinetic energy will become half of initial kinetic energy. Solution: Let this situation occur at distance x. Then apply conservation of energy: (KE + PE)surface = (KE + PE)distance 1/2m(GM/R) - GMm/R = 1/2 * 1/2m(GM/R) - GMm/x => 1/x = 1/R + 1/8R - 1/4R = 1/R[1 + 1/8 - 1/4] = 7/8R => x = 8R/7 Q-11 A block of mass 'm' attached to a spring of spring constant 'k'. The spring is pulled by a horizontal force of constant magnitude F. Potential energy of spring at the moment when block momentarily comes to rest is Solution: The block momentarily comes to rest at extreme point where the elongation in the spring is (2F/k) [from question 6] ∴ PE = ½kx² = ½k(2F/k)² = 2F²/k