Question

Elasticity, Hooke's Law, Periodic and Simple Harmonic Motion: A 1.6 kg mass is hung from a 25 cm long coiled spring. The the total surface area of the spring is 3.20 10-4 m? (Assume no friction or air resistance, and a "mass-less" spring) Part 1. If the spring stretches 16 cm when the mass is added, what is the Spring Constant (k)? Give your answer to 2 significant digits. kg/s2 Part 2. What is the Stress on the spring? Give your answer to nearest thousand Pa. Pa Part 3. What is the Strain on the spring? Give your answer to 2 significant digits. Part 4. If you displace the spring an additional 14 cm. What is the maximum velocity during the motion? (Hint: think about it in terms of Kinetic and Potential Energy, also the "max displacement A = 14 m) Give your answer to 3 significant digits. m/s Part 5. What is the Period of vibration? Give your answers to 3 significant digits. S

Answer 1

K and somn lit consider a spring of Spring Constant mars m, = 1.6 kg. The natural Spring is l. -YO length of l= 25cm. Natural length. a) when extension from epailton (x=0) 2 mi equilibrium is Ax = 0.16m then. Kax = a = mg K = 1•6x9.fr 0:16. 2 K=98 N/m. or K = 98-00 N/m b) Stress on Spring is stress = mig where A= 3.2x10-4 m² А Istress =. 4 • gox10 4 Pascal. Stress = 1-6x3-8 3.2 X10-9 6 strain = AR 0.16 strain = 0.64 . a 0.25 d) if we displace by amount 4y = 0.14 m. , it will execute Simple harmonic motion, and we know , at equilibrium Velocity is maximum. Total energy at equilibrium is E, = 1 m, u 2 + 1 kox? lotal energy at extreme point is E₂ tok (ol+0.14)2 2 E, = £m, 0.4+, x 96% 0.162 '1 2 E = 1 gb x 0.32 2 2 Imavza (98) (0-32-0.163) m, uz=6.3112 or v² = 6-3112 1.6 = 3.9445 v=1.986 mla

so = kx or w km, Since mans execute Simple harmonic motion with frequency w mw²x = 7.826 w = 7.326 radio. then time period of oscillation is T= 2n =) T=0.8028 S 98 106 w