1)I attach a 2.0-kg block to a spring that obeys Hookes Law and supply 16 J of e
ID: 1684230 • Letter: 1
Question
1)I attach a 2.0-kg block to a spring that obeys Hookes Law and supply 16 J of energy to stretch the spring. I release the block; it oscillates with period 0.30s. The amplitude is A) 38cm B)19cm C)9.5cm D)4.3cm 2) The motion of a piston in an automobile engine is nearly simple harmonic. If the 1-kg piston travels back and forth over a total distance of 10.0cm, what is its maximum speed when the engine is running at 3000rpm? A)31.4m/s B)15.7m/s C)7.85m/s D)3.93m/s 3)The position of a 0.64-kg mass undergoing simple harmonic motion is given by x=(0.160m) cos(pt/16). What is the maximum net force on the mass as it oscillates? A)3.9x10^-3N B)9.9x10^-3N C)1.3x10^-3N D)6.3N 4)A simple pendulum has a period of 2.0s. What is the pendulum length (g=9.8m/s^2) A) 0.36m B)0.78m C)0.99m D)2.4m5) A wave is traveling in a string at 60 m/s. When the tension is then increased 20%, what will be the resulting wave speed? A) also 60m/s B) 66 m/s C)72m/s D)55m/s
1)I attach a 2.0-kg block to a spring that obeys Hookes Law and supply 16 J of energy to stretch the spring. I release the block; it oscillates with period 0.30s. The amplitude is A) 38cm B)19cm C)9.5cm D)4.3cm 2) The motion of a piston in an automobile engine is nearly simple harmonic. If the 1-kg piston travels back and forth over a total distance of 10.0cm, what is its maximum speed when the engine is running at 3000rpm? A)31.4m/s B)15.7m/s C)7.85m/s D)3.93m/s 3)The position of a 0.64-kg mass undergoing simple harmonic motion is given by x=(0.160m) cos(pt/16). What is the maximum net force on the mass as it oscillates? A)3.9x10^-3N B)9.9x10^-3N C)1.3x10^-3N D)6.3N 4)A simple pendulum has a period of 2.0s. What is the pendulum length (g=9.8m/s^2) A) 0.36m B)0.78m C)0.99m D)2.4m
5) A wave is traveling in a string at 60 m/s. When the tension is then increased 20%, what will be the resulting wave speed? A) also 60m/s B) 66 m/s C)72m/s D)55m/s
Explanation / Answer
1. 1/2kx^2 = 16 J ? = v(k/m)k = ?^2m, and ? = 2p/T k = (2p/.30)^2 *2 = 877.3 N/m x = v(2*16 / 877.3) = .19 m = 19 cm (Choice B)
2. amplitude: .100 m/4 = .025 m ? = 3000 rev/min *2p rad/rev* 1 min/60 sec = 314.2 rad/s x = Acos(?t) and v = -?Asin(?t) maximum speed = ?A = 314.2 rad/s * .025 m = 7.85 m/s (Choice C)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.