PHYSICS - CLUTCH CH 15: PERIODIC MOTION (NEW)

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2 CONCEPT: Hooke s Law & Springs When you push/pull against a spring (FA), spring pushes back in the direction. (Action-Reaction!) Fs = FA = Ex. 1: You push on a spring with a force of 120N. The spring constant k is 20. How much does it compress? x = D - Relaxed position (x = ) - NOT the spring s length (x = ) k = spring s force constant - Measures how the spring is. - Higher k to deform Ex. 2: How much force is required to pull a spring of length 10m out to 16m, if the spring constant k is 40N/m? - Ex. 1: x = k = F = - Ex. 2: x = k = F = - Units of k: FS = R force, always opposes deformation EXAMPLE 3: A 1.0 m-long spring is laid horizontally with one of its ends fixed. When you pull on it with 50 N, it stretches to 1.2 m. (a) What is the spring s force constant? (b) How much force is needed to compress it to 0.7 m? Page 2

3 PRACTICE: A 1.0 m-long spring is laid horizontally with one of its ends fixed. When you pull on it with 50 N, it stretches to 1.2 m. (a) What is the spring s force constant? (b) How much force is needed to compress it to 0.7 m? Page 3

4 CONCEPT: Spring Forces If you attach a mass to a spring (mass-spring system) and release, the force pulls it back to equilibrium. - The m always refers to the mass of the (springs are always massless!) - Compressed: ΣF = ma m a = - Released: ΣF = ma m a = EXAMPLE 1: A 0.60-kg block attached to a spring with force constant 15 N/m. The block is released from rest when the spring is stretched 0.2 m to the right. At the instant the block is released, find (a) the force on the block and (b) its acceleration, assuming right is positive. m PRACTICE: You push a 3-kg mass against a spring and release it from rest. Its maximum acceleration is 10m/s 2 when pushed back 0.5m. What is the (a) spring constant and (b) restoring force at this point? Page 4

5 CONCEPT: Intro to Simple Harmonic Motion The most common type of Simple Harmonic Motion (aka Oscillation) is the mass-spring system. (EQ) Amplitude A: - displacement, x - always the. x = = v = x = v = x = = v = Period T [seconds/cycle] - Time for one cycle. Frequency f = 1/T [cycles/second] F = a = F = a = F = a = Angular frequency ω [rad/second] - ω = = EXAMPLE 1: A mass on a spring is pulled 1m away from its equilibrium position, then released from rest. The mass takes 2s to reach maximum displacement on the other side. Calculate the (a) amplitude, (b) period, (c) angular frequency of the motion. Page 5

6 PRACTICE: A mass-spring system with an angular frequency ω = 8π rad/s oscillates back and forth. (a) Assuming it starts from rest, how much time passes before the mass has a speed of 0 again? (b) How many full cycles does the system complete in 60s? Mass-Spring SHM Equations F S = F A = kx a = k m x ω = 2πf = 2π T = k m t [time] N [cycles] = = t f T [Period] Page 6

7 CONCEPT: Equations for Simple Harmonic Motion In Simple Harmonic Motion, acceleration NOT constant kinematic equations? Old Equations x (position) F s = k x Fmax = a = k x m amax = New Equations t (time) x(t) = + A cos(ωt) xmax = v(t) = Aω sin(ωt) vmax = a(t) = Aω 2 cos(ωt) amax = - Calculator must be in. Combining amax(x) and amax(t) ω = 2πf = 2π T = EXAMPLE 1: A 4-kg mass is attached to a spring where k = 200[N/m]. The mass is pulled 2m and released from rest. Find the (a) angular frequency, (b) velocity 0.5s after release, (c) acceleration when x = 0.5m, and (d) the period of oscillation. Page 7

8 PRACTICE: A 4-kg mass on a spring is released 5 m away from equilibrium position and takes 1.5 s to reach its equilibrium position. (a) Find the spring s force constant. (b) Find the object s max speed. -A 0 +A Mass-Spring SHM Equations F S = F A = kx Fmax = ±ka a = k m x amax = ± k m A x(t) = + A cos(ωt) xmax = ±A v(t) = Aω sin(ωt) vmax = ±Aω a(t) = Aω 2 cos(ωt) amax = ±Aω 2 ω = 2πf = 2π T = k m t [time] N [cycles] = = t f T [Period] EXAMPLE: A 4-kg mass is attached to a horizontal spring and oscillates at 2 Hz. If mass is moving with 10 m/s when it crosses its equilibrium position, (a) how long does it take to get from equilibrium to its max distance? Find the (b) amplitude and (c) maximum acceleration. -A 0 +A Mass-Spring SHM Equations F S = F A = kx Fmax = ±ka a = k m x amax = ± k m A x(t) = + A cos(ωt) xmax = ±A v(t) = Aω sin(ωt) vmax = ±Aω a(t) = Aω 2 cos(ωt) amax = ±Aω 2 ω = 2πf = 2π T = k m t [time] N [cycles] = = t f T [Period] Page 8

9 PRACTICE: What is the equation for the position of a mass moving on the end of a spring which is stretched 8.8cm from equilibrium and then released from rest, and whose period is 0.66s? What will be the object s position after 1.4s? -A 0 +A Mass-Spring SHM Equations F S = F A = kx Fmax = ±ka a = k m x amax = ± k m A x(t) = + A cos(ωt) xmax = ±A v(t) = Aω sin(ωt) vmax = ±Aω a(t) = Aω 2 cos(ωt) amax = ±Aω 2 ω = 2πf = 2π T = k m t [time] N [cycles] = = t f T [Period] EXAMPLE: The velocity of a particle on a spring is given by the equation v(t) = sin (3π t). Determine the (a) frequency of motion, (b) amplitude, and (c) velocity of the particle at t = 0.5s. -A 0 +A Mass-Spring SHM Equations F S = F A = kx Fmax = ±ka a = k m x amax = ± k m A x(t) = + A cos(ωt) xmax = ±A v(t) = Aω sin(ωt) vmax = ±Aω a(t) = Aω 2 cos(ωt) amax = ±Aω 2 ω = 2πf = 2π T = k m t [time] N [cycles] = = t f T [Period] Page 9

10 CONCEPT: Energy in Simple Harmonic Motion At any point of SHM, the mass-spring system may have 2 types of Energy: +. - Wnc =, so the Mechanical Energy (M.E.) is. (EQ) m Energy Conservation always compare energies at 2 special points: Any other Point: Amplitude: x = Elastic Energy (UA) = 1 2 kx2 = Kinetic Energy (KA) = 1 2 mv2 = Total M.E. = Equilibrium: x = Elastic Energy (U0) = 1 2 kx2 = Kinetic Energy (K0) = 1 2 mv2 = Total M.E. = x = Elastic Energy (UP) = 1 2 kx2 = Kinetic Energy (KP) = 1 2 mv2 = Total M.E. = Comparing all these energies at different points: U A = K 0 = U P + K P = = + v(x) = (Energy Conservation for Springs) EXAMPLE 1: A 5 kg mass oscillates on a horizontal spring with k = 30[N/m] and an amplitude of 0.4 m. Find its (a) max speed, (b) speed when it is at -0.2 m, and (c) the total mechanical energy of the system. Page 10

11 EXAMPLE: A 0.25-kg mass oscillates on a spring with a period of 3.2s. At x=0.4m, it is observed to have a speed of 5m/s. What is the system s (a) Amplitude and (b) total mechanical energy? -A 0 +A Mass-Spring SHM Equations F S = F A = kx Fmax = ±ka a = k m x amax = ± k m A x(t) = + A cos(ωt) xmax = ±A v(t) = Aω sin(ωt) vmax = ±Aω a(t) = Aω 2 cos(ωt) amax = ±Aω 2 ω = 2πf = 2π T = k m t [time] N [cycles] = = t f T [Period] M. E. = 1 2 ka2 = 1 2 mv 2 max = 1 2 kx p mv p 2 v(x) = ω A 2 x 2 PRACTICE: A block of mass kg is attached to a spring. At x = m, its acceleration is a x = m/s 2 and its velocity is v x = 4.00 m/s. What are the system s (a) force constant k and (b) amplitude of motion? -A 0 +A Mass-Spring SHM Equations F S = F A = kx Fmax = ±ka a = k m x amax = ± k m A x(t) = + A cos(ωt) xmax = ±A v(t) = Aω sin(ωt) vmax = ±Aω a(t) = Aω 2 cos(ωt) amax = ±Aω 2 ω = 2πf = 2π T = k m t [time] N [cycles] = = t f T [Period] M. E. = 1 2 ka2 = 1 2 mv 2 max = 1 2 kx p mv p 2 v(x) = ω A 2 x 2 Page 11

12 EXAMPLE: You increase the amplitude of oscillation of a mass vibrating on a spring. Which statements are correct? (a) Period of oscillation increases (b) Maximum acceleration increases (c) Maximum speed increases (c) Max Kinetic Energy increases (d) Max Potential Energy increases (e) Max Total Energy increases -A 0 +A Mass-Spring SHM Equations F S = F A = kx Fmax = ±ka a = k m x amax = ± k m A x(t) = + A cos(ωt) xmax = ±A v(t) = Aω sin(ωt) vmax = ±Aω a(t) = Aω 2 cos(ωt) amax = ±Aω 2 ω = 2πf = 2π T = k m t [time] N [cycles] = = t f T [Period] M. E. = 1 2 ka2 = 1 2 mv 2 max = 1 2 kx p mv p 2 v(x) = ω A 2 x 2 Page 12

13 CONCEPT: Vertical Oscillations Vertical mass-spring systems very similar to Horizontal, except: - Horizontal Equilibrium at relaxed position (x = 0) - Vertical Equilibrium where forces. EQ EXAMPLE: You hang a 0.5m spring from the ceiling. When you attach a 5kg mass to it, it stretches by 0.2m. You pull the mass-spring system down an additional 0.3m and release. Find (a) the spring constant. (b) At its maximum height, how far from the ceiling is the block? Page 13

14 PRACTICE: A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed? -A 0 +A Mass-Spring SHM Equations F = F = kx Fmax = ±ka a = x amax = ± A x(t) = + A cos(ωt) xmax = ±A v(t) = Aω sin(ωt) vmax = ±Aω a(t) = Aω cos(ωt) amax = ±Aω ω = 2π = 2π T = k m [cycles] = [time] [Period] =.. = 1 2 = 1 2 = ( ) = EXAMPLE: An elastic cord is 65 cm long when a weight of 75 N hangs from it but is 85 cm long when a weight of 180 N hangs from it. What is the spring constant k of this elastic cord? PRACTICE: A chair of mass 30 kg on top of a spring oscillates with a period of 2s. (a) Find the spring s force constant. You place an object on top of the chair, and it now oscillates with a period of 3s. (b) Find the object s mass. Page 14

15 CONCEPT: Simple Pendulums Just like mass-spring systems, pendulums also display Simple Harmonic Motion. Mass-Spring (EQ) F = = 2 = = m a = Pendulum F = = 2 = = m a = - Make sure θ and calculator is in RADIANS! For SHM, restoring force must be proportional to deformation/distance. - For small angles,. Restoring Force: Example 1: You pull a 0.250m long pendulum with a hanging mass of 4kg to the side by 3.50 and release. Find the (a) restoring force, (b) period of oscillation, (c) time it takes for the mass to reach its maximum speed. m Page 15

16 PRACTICE: A pendulum makes 120 complete oscillations in 3.00 minutes. Find (a) the period of oscillation and (b) the length of the pendulum. Pendulum SHM Equations F = mgθ a = gθ = ω = 2π = 2π T = [cycles] = [time] [Period] = EXAMPLE: After landing on an unfamiliar planet, an astronaut constructs a simple pendulum of length 3m and mass 4kg. The astronaut releases the pendulum from 10 degrees with the vertical, and clocks one full cycle at 2s. Calculate the acceleration due to gravity at the surface of this planet. Pendulum SHM Equations F = mgθ a = gθ = ω = 2π = 2π T = [cycles] = [time] [Period] = Page 16

17 CONCEPT: Pendulum SHM Equations -A 0 +A Mass-Springs x = deformation from EQ xmax = A vmax = Aω amax = Aω 2 Pendulums At any point: x = xmax = = vmax = Aω = amax = Aω 2 = ω =2πf = 2π/T= ω =2πf = 2π/T= m If asked for given t ( ) = ( ) EXAMPLE: A 500-g mass hangs from a 40-cm-long string. The object has a speed of 0.25m/s as it passes through its lowest point. What maximum angle (in degrees) does the pendulum reach? Page 17

18 PRACTICE: A simple pendulum is 0.30m long. At t = 0, it is released from rest, starting at an angle of 13. What will be the angular position (in degrees) of the pendulum at t = 0.35s? Pendulum SHM Equations F = F = mgθ a = gθ = ω = 2π = 2π T = [cycles] = [time] [Period] = xmax = A = Lθ ( ) = ( ) vmax = Aω = Lθmaxω amax = Aω 2 = Lθmaxω 2 EXAMPLE: A 100g mass on a 1.0m-long string is pulled 7.0 to one side and released. How long does it take to reach 4.0 on the opposite side? Pendulum SHM Equations F = F = mgθ a = gθ = ω = 2π = 2π T = [cycles] = [time] [Period] = xmax = A = Lθ ( ) = ( ) vmax = Aω = Lθmaxω amax = Aω 2 = Lθmaxω 2 Page 18

19 CONCEPT: Energy in Simple Pendulums Just like mass-spring systems, energy in pendulums 2 types: and. m For any θ, height h = Amplitude: Equilibrium: Any other Point: θ = θ = θ = Grav. Potential = h = 0 / max Grav. Potential = h = 0 / max Grav. Potential = h Kinetic Energy = = 0 / max Kinetic Energy = = 0 / max Kinetic Energy = Total M.E. = Total M.E. = Total M.E. = + M.E. = = = + (Energy Conservation for Pendulums) EXAMPLE: A mass m is attached to a pendulum of length L. It is pulled up an angle θ and released. Using energy conservation, derive an expression for the maximum speed this mass experiences. Page 19

20 EXAMPLE: A mass of kg hangs from a 2m pendulum. At the moment when it makes a 5 with the vertical, it has a speed of 1.5 m/s. What is the maximum height the pendulum will reach? Pendulum SHM Equations F = F = mgθ a = gθ = ω = 2π = 2π T = [cycles] = [time] [Period] = xmax = A = Lθ ( ) = ( ) vmax = Aω = Lθmaxω amax = Aω 2 = Lθmaxω 2 PRACTICE: A mass swinging at the end of a pendulum has a speed of 1.32m/s at the bottom of its swing. At the top of its swing, it makes a 9 with the vertical. What is the length of the pendulum? Pendulum SHM Equations F = mgθ a = gθ = ω = 2π = 2π T = [cycles] = [time] [Period] = ( ) = ( ) A = Lθ = 2 (1 cos ) vmax = Aω h = (1 ).. = h = 1 2 = h Page 20