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1. A mass m at the end of a spring vibrates with a frequency of 0.88 Hz. When an additional 600 g mass is added to m, the frequency is 0.60 Hz. What is the value of m? ω= √ √ k 1 k → f= m 2π m f2 (0.6 Hz) m m = → = f1 m2 m+0.600 kg (0.88 Hz) √ √ → m=0.52 kg. 2. It takes a force of 80.0 N to compress the spring of a popgun 0.20 m to load a 0.15 kg ball. With what speed will the ball leave the gun? k =F / Δ x=(80.0N)/(0.20m)=400 N /m The ball will leave at the equilibrium position, where the kinetic energy is maximum. The maximum potential energy equals the maximum kinetic energy, therefore we have: 1 1 KE Max = m v 20 =PE Max = kA 2 ; 2 2 1 1 (0.150 kg) v 20= (400 N / m)(0.20 m)2 2 2 → v 0=10.3 m/ s. 3. A 0.5 kg mass vibrates according to the equation x=0.45cos(8.4t), where x is in meters and t is in seconds. Determine A=0.45 m a) the amplitude b) the frequency 2 π f =8.4 s−1 → f =1.34 Hz c) total energy v 0=ω A=(8.4 s−1)(0.45m)=3.78 m/ s 1 1 E= KE Max = mv 20 = (0.5 kg)(3.78 m/ s)2=3.6 J. 2 2 d) potential energy when x=0.3 m. 1 2 1 1 2 2 −1 2 2 PE= k x = (m ω ) x = (0.5 kg)(8.4 s ) (0.3 m) =1.6 J. 2 2 2