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This is a description of the motion of a mass on a spring as a cosine function. It includes information about the frequency of the oscillation, the object's speed at different positions, and the phase angles at different time intervals.
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A mass on a spring oscillates (–10.0 cm ≤ x ≤ 10.0 cm) as shown in this position vs. time plot. At the moment indicated by P, t = 3.00 s, and x = 4.00 cm. Write the description of the mass’s motion as a cosine function. • x(t) = 10 cm[cos(0.386t)] • x(t) = 10 cm[cos(3.00t + 0.400)] • x(t) = 10 cm[cos(22.1t)] • Not enough information. • None of the above. Oregon State University PH 212, Class #21
Frequency of an Object in Simple Harmonic Motion If no external work is done on the spring-mass system, its total mechanical energy remains constant throughout its oscillation: Emech.total = KT + Uspr = (1/2)mv2 + (1/2)kx2 And we know something in particular about Emech.total at certain points in the oscillation cycle: Where all the energy is in the form of KT (at x = 0), Emech.total = (1/2)mvmax2 = (1/2)m[ωA]2 Where all the energy is in the form of Uspr (at x = ±A), Emech.total = (1/2)kA2 Thus: (1/2)kA2 = (1/2)m[ωA]2 Or: ω = √(k/m) Oregon State University PH 212, Class #21
An object moves horizontally with simple harmonic motion. At which position is the object’s speed equal to half of its maximum speed? (A = amplitude) • x = A/2 • x = (√2)A/2 • x = (√3)A/2 • Not enough information. • None of the above. Oregon State University PH 212, Class #21
A mass on a spring oscillates horizontally with a period of 1.50 s. At t = 0 s, the mass is located 5.00 cm to the left of the equilibrium position, and it is moving to the right at a speed of 36.3 cm/s. Assuming that we are describing the motion with a cosine function, find the initial phase angle and the phase angle at t = 1.00 s. That is: Find f(0), which we call f0; also, find f(1). • f0 = 1.05 rad f(1) = 5.24 rad • f0 = –1.05 rad f(1) = 3.14 rad • f0 = 4.19 rad f(1) = 8.38 rad • Not enough information. • None of the above. Oregon State University PH 212, Class #21
A mass on a spring oscillates horizontally with a period of 1.50 s. At t = 0 s, the mass is located 5.00 cm to the left of the equilibrium position, and it is moving to the right at a speed of 36.3 cm/s. Assuming that we are describing the motion with a cosine function, find the initial phase angle and the phase angle at t = 1.00 s. That is: Find f(0), which we call f0; also, find f(1). • f0 = 1.05 rad f(1) = 5.24 rad • f0 = –1.05 rad f(1) = 3.14 rad • f0 = 4.19 rad f(1) = 8.38 rad • Not enough information. • None of the above. Oregon State University PH 212, Class #21
Four common SH oscillators All from the same fundamental equation: acceleration = –2(position) Horizontal mass-spring: = √(k/m) Vertical mass-spring: = √(k/m) Simple pendulum: = √(g/L) Physical pendulum: = √(Mglcm.pivot/Ipivot) (See also page 413 for another good summary.) Oregon State University PH 212, Class #21
A uniform steel bar swings freely from a pivot at one end. The period of its oscillation is 1.2 s. How long is the bar?[Reminders: Irod.end = (1/3)ML2 and g = 9.80 m/s2) • L = 0.536 m • L = 1.07 m • L = 2.14 m • Not enough information. • None of the above. (Follow-up: How long would the arm of a simple pendulum need to be in order to match the period of the above physical pendulum?) Oregon State University PH 212, Class #21
A uniform steel bar swings freely from a pivot at one end. The period of its oscillation is 1.2 s. How long is the bar?[Reminders: Irod.end = (1/3)ML2 and g = 9.80 m/s2) • L = 0.536 m • L = 1.07 m • L = 2.14 m • Not enough information. • None of the above. (How long would the arm of a simple pendulum need to be in order to match the period of the above physical pendulum? 0.357 m) Oregon State University PH 212, Class #21
