This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Week 1 Sample Problems The difficulty of each problem is listed immediately after the problem number. Problem 1 (medium): A mass m = 1 kg is connected to a spring with spring constant k = 5 N/m, and rests on a flat frictionless surface. The mass is compressed into the spring, to the right , by 3 m, and at t = 0, is released from rest at this location. Define right as the positive direction, and left as the negative direction. 1: What is the mass’s position as a function of time? 2: What is the maximum (positive) velocity of the mass? At what time does this first occur? 3: What is the maximum (positive) acceleration of the mass? At what time does this first occur? Solution: We define right as positive, and left as negative. The position of the mass as a function of time is given by x ( t ) = x m cos( ωt ) (1) where x m = 3 m ω = r k m = r 5 1 = 2 . 24 rad / s (2) The velocity and acceleration of the mass, as a function of time, are given by v ( t ) = dx dt = ωx m sin( ωt ) a ( t ) = dv dt = ω 2 x m cos( ωt ) (3) The maximum positive velocity of the mass is v m = ωx m = (2 . 24)(3) = 6 . 71 m / s (4) This velocity is first reached at a time t 1 , when v m sin( ωt 1 ) = v m sin( ωt 1 ) = 1 ωt 1 = 3 π 2 t 1 = 3 π 2 ω = 3 π 2 √ 5 = 2 . 11 s (5) 1 The maximum positive acceleration of the mass is a m = ω 2 x m = (5)(3) = 15 m / s 2 (6) This acceleration is first reached at a time t 2 , when a m cos( ωt 2 ) = a m cos( ωt 2 ) = 1 ωt 2 = π t 2 = π ω = π √ 5 = 1 . 40 s (7) Problem 2 (mediumhard): A mass m = 2 kg is connected to a spring, and rests on a flat frictionless surface. The mass oscillates horizontally in simple harmonic motion. At t = 3 s, the mass has its maximum velocity of v m = 5 m/s, directed to the left. The amplitude of the oscillations is x m = 10 m. 1: What is the spring constant k of the spring? 2: What is the position of the mass as a function of time? 3: At a certain time, the mass’s potential energy is 1/6 of its kinetic energy, and the mass is to the right of its equilibrium position but moving toward equilibrium. What are the positions, velocity, and acceleration of the mass at this time? 4: Find all times at which the mass’s acceleration is 1 m/s 2 , directed to the left. What is the earliest positive time ( t > 0) at which this occurs? Solution: We determine the frequency ω of the motion using the amplitudes of the velocity and position: ω = v m x m = 5 10 = 0 . 5 rad / s (1) The spring constant k of the spring is therefore k = mω 2 = (2)(0 . 5) 2 = 0 . 5 N / m (2) We define right as positive, and left as negative. The mass has maximum negative velocity at t = 3 s. Therefore, its velocity as a function of time is v ( t ) = v m cos( ω ( t 3)) = 5cos(0 . 5( t 3)) m / s (3) The position of the mass as a function of time is therefore 2 x ( t ) = Z v ( t ) dt = v m ω sin( ω ( t 3)) = x m sin( ω ( t 3)) = 10sin(0 . 5( t 3)) m (4) The position of equilibrium is defined to be x = 0, which is satisfied by the above equation.= 0, which is satisfied by the above equation....
View
Full Document
 Spring '09
 WAUNG
 Mass, Simple Harmonic Motion, Velocity, Ycm

Click to edit the document details