RECALL: Angular Displacement & Angular Velocity n Think about spinning a ball on a string or a ball on a turn-table in terms of SHM and UCM: n If you look at the ball from the side, its motion could be plotted as a sine wave. n Δθ is the angle between position 1 and 2. n Δθ is the amount of rotation—measured in degrees or radians (we’ll use radians since they are the SI unit for θ). 2 θ r 1 For an object rotating about a fixed axis, the angular displacement is the angle Δθ swept out by a line passing through any point on the body and intersecting the axis of rotation perpendicularly. By convention, the angular displacement is positive if it is counterclockwise. Angular Displacement, θ n For an object rotating, angular displacement is found by: n θ (in radians) = Arc length = s radius r n Δθ = θf – θi where θi is sometimes zero. If determining Δθ for a complete circle, or one complete cycle: n Arc length = 2πr (for one revolution) n Δθ = (arc length) / r n = (2πr) / r = 2π radians n ∴ 1 revolution = 2π radians = 360o n So, 1 radian = 360o / 2π = 57.3o Once around the circular path is 360-degrees or 2π-radians! Angular Velocity n Angular Velocity = the “turning rate”—expressed in units of revolutions per second, or metric: radians per second. n Angular velocity = angular displacement / elapsed time n ω = Δθ/Δt units: radians/second or rad/s n For one cycle: Δt = T = period = time for one cycle n And, Δθ = (2πr)/r = 2π radians = angular displacement for one cycle n ∴ ω = Δθ/Δt = 2π radians / T = 2π / T n ω = 2π / T n And since T=1/f , where f = frequency n ω = 2π / (1/f) = 2πf n ω = 2πf ∴ Angular velocity is sometimes called angular frequency! Important to realize… n Angular velocity, ω, or the “turning rate” is the same everywhere on the rotating body/object, b/c Δθ/Δt is the same. n But, instantaneous velocity, the velocity tangent to the circular path—called tangential velocity, vT– is greater for points (places) farther from the axis. n The magnitude of the tangential velocity is referred to as the tangential speed. n vT= r ω (new) SHM & the Reference Circle n The ball mounted on the turntable moves in uniform circular motion on a path known as the reference circle, and its shadow, projected on a moving strip of film, executes simple harmonic motion. Determining displacement, x, for an object experiencing simple harmonic motion n Reference Circle of radius A n Ball starts on x-axis at x=+A and moves through an angle θ in a time t. n Since moving in UCM, angular speed ω is constant. n Recall: θ = ωt n The displacement x of the shadow is the x-component of the radius A: n Cosine θ = x/A, or n X = A cos θ = A cos (ωt) (And: ω = 2πf) n The shadow of the ball oscillates between values of x=+A and x=-A n The radius A of the reference circle is the amplitude of the SHM Determining velocity, v, for an object experiencing simple harmonic motion n vT= tangential velocity = r ω n The velocity v of the shadow is the x-component of the tangential velocity n Sine θ = -v/vT n Negative b/c v is pointing in the direction of the negative x-axis. n Recall, θ = ωt and vT=rω n And since r = A, vT= Aω n ∴the velocity of an object in SHM is given by: n v= -vT sin θ n v= -Aω sin θ = -A ω sin (ωt) n In general: (ω = 2πf) n v= +/- A ω sin θ n v= +/- A ω sin (ωt) n This velocity is not constant, but varies between maximum and minimum values as time passes. n When the shadow changes direction at either end of the oscillatory motion, the velocity is momentarily zero. n When the shadow passes through the x=0m position, the velocity has a maximum amount of v=Aω, since the sine of an angle is between +1 and -1. n vmax = A ω n (ω = 2πf) n Both amplitude and angular frequency determine the maximum velocity. n A person bounces up and down on a trampoline, while always staying in contact with it. n The motion is simple harmonic motion, and it takes 1.90 seconds to complete one cycle. n The height of each bounce above the equilibrium position is 45.0 cm. n Determine the amplitude, and n determine the angular frequency of the motion. n What is the maximum speed attained by the person? The solution… n The amplitude of simple harmonic motion is the distance from the equilibrium position to the point of maximum height. n Since the distance from the equilibrium position to the point of maximum height is the amplitude of the motion… n A = 45.0 cm = 0.45 m n The angular frequency is found with ω = 2π / T n ω = 2π / T = 2π / 1.90 seconds n ω = 3.31 rad/second n The maximum speed attained by the person on the trampoline depends on the amplitude and the angular frequency of the motion: n vmax = A ω = (0.45m)(3.31 rad/s) = 1.49 m/s= vmax In SHM, the velocity is not constant, ∴the object experiencing SHM is accelerating… n An object moving in a circle experiences UCM; 2 n ∴ ac = v /r n So, for an object experiencing SHM: n (tangential) Speed = v = Δd/Δt = 2 π r/T= 2 π r·f n v=2 π f·r = (2 π f)r = ω⋅r = r⋅ω n ∴v = r⋅ω (where v is the tangential speed) n Putting it all together… 2 2 2 2 n ac = v /r = (r⋅ω) /r =r⋅ω = A⋅ω (since r = A for SHM) 2 n ac = A⋅ω Like velocity, acceleration is not constant as time passes. n a is the x-component of ac n Cosine θ = -a/ac n Negative b/c pointing in “–x direction” n a = -ac cos θ 2 n And ac = Aω n a = -Aω2 cos θ n And θ = ωt n a = -Aω2 cos (ωt) n In general: n a = +/-Aω2 cos (ωt) n And (ω = 2πf) 2 n amax = Aω Slide 14-22 n Check this out! n www.acoustics.salford.ac.uk/feschools/waves/ shm.php Frequency of Vibration n For an object on a spring moving back and forth, ignoring friction, where the only force acting on the spring causing it to vibrate is the restoring force of the spring: n ΣF = Frestoring = -kx = ma n -kx = ma n -k [A cos (ωt)]= m[-Aω2 cos (ωt)] n -k [A]= m[-Aω2] n k = mω2 n ω2 = k/m n ω = √(k/m) n Recall ω=2πf n ω = 2πf = √(k/m) n for a spring w/mass m on it n where ω is in radians/second The frequency of oscillation depends on physical properties of the oscillator; it does not depend on the amplitude of the oscillation. nω = 2πf = √(k/m) n for a spring w/mass m on it n where ω is in radians/second Rearranged: Slide 14-14 n The diaphragm of a loudspeaker moves back and forth in simple harmonic motion to create sound. The frequency of the motion is 1.0 kHz and the amplitude is 0.20 mm. n A) What is the maximum speed of the diaphragm? n B) Where in the motion does this maximum speed occur? n C) What is the maximum acceleration of the diaphragm? n D) Where in the motion does this maximum acceleration occur? Solution: nω = 2πf nv max = A ω nv max=(0.0002m)2π (1000Hz) n A) vmax = 1.3 m/s n B) The speed of the diaphragm is momentarily zero at the two ends of its motion, x = +A and x = -A… its maximum speed occurs midway between these two points, or at x=0m. 2 2 2 n C) amax = Aω = (0.0002m)[2π (1000Hz)] =amax = 7900 m/s n D) The maximum acceleration occurs when the force acting on the diaphragm is a maximum. The maximum force acts when the diaphragm is at the ends of the path, where the displacement is the greatest… ∴the maximum acceleration is at x=+A and x=-A. Auto Suspensions The shock absorbers in the suspension system of a car are in such bad shape that they have no effect on the behavior of the springs attached to the axles. Each of the identical springs attached to the front axle supports 320 kg. A person pushes down on the middle of the front end of the car and notices that it vibrates through five cycles in 3.0 seconds. Find the spring constant of either spring. n ω=2πf = √(k/m) n 2πf = √(k/m) n (2πf)2 = k/m n m⋅4π2 ⋅ f2 = k n (320 kg)(4π2)(5cycles/3sec)2 = k n k = 35091.9 N/m A ball on a spring is pulled down and then released. Its subsequent motion appears as follows: 1. At which of the above times is the displacement zero? 2. At which of the above times is the velocity zero? 3. At which of the above times is the acceleration zero? 4. At which of the above times is the kinetic energy a maximum? 5. At which of the above times is the potential energy a maximum? 6.