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**Harmonic Motion and Waves**

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**Simple Harmonic Motion(SHM)**

Vibration (oscillation) Equilibrium position – position of the natural length of a spring Amplitude – maximum displacement

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Period and Frequency Period (T) – Time for one complete cycle (back to starting point) Frequency (Hz) – Cycles per second F = 1 T = 1 T f

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Period and Frequency A radio station has a frequency of M Hz. What is the period of the wave? 103.1 M Hz 1X106 Hz = X 108 Hz 1M Hz T = 1/f = 1/(1.031 X 108 Hz) = X 10-9 s

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**Hooke’s Law F = -kx F = weight of an object k = spring constant (N/m)**

x = displacement when the object is placed on the spring

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Hooke’s Law: Example 1 What is the spring constant if a kg mass causes the spring to stretch 6.0 cm? (ANS: 16 N/m)

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Special Note: If a spring has a mass on it, and then is stretched further, equilibrium position is the starting length (with the mass on it)

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Hooke’s Law: Example 2 A family of four has a combined mass of 200 kg. When they step in their 1200 kg car, the shocks compress 3.0 cm. What is the spring constant of the shocks? F = -kx k = -F/x k = -(200 kg)(9.8 m/s2)/(-0.03 m) k = 6.5 X 104 N/m (note that we did not include the mass of the car)

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Hooke’s Law: Example 2a How far will the car lower if a 300 kg family borrows the car? F = -kx x = -F/k x = -(300 kg)(9.8 m/s2)/ 6.5 X 104 N/m x = 4.5 X 10-2 m = 4.5 cm

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**Forces on a Spring Extreme Position (Amplitude) Equilibrium position**

Force at maximum Velocity = 0 Equilibrium position Force = 0 Velocity at maximum

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**Energy and Springs KE = ½ mv2 PE = ½ kx2 Maximum PE = ½ kA2**

Law of conservation of Energy ½ kA2 = ½ mv2+ ½ kx2

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All PE All KE Some KE and Some PE

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**Spring Energy: Example 1**

A 0.50 kg mass is connected to a light spring with a spring constant of 20 N/m. Calculate the total energy if the amplitude is 3.0 cm. Maximum PE = ½ kA2 Maximum PE = ½ (20 N/m)(0.030 m2) Maximum PE = 9 X 10-3 Nm (J)

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**Spring Energy: Example 1a**

What is the maximum speed of the mass? ½ kA2 = ½ mv2+ ½ kx2 ½ kA2 = ½ mv2 (x=0 at the origin) 9 X 10-3 J = ½ (0.50 kg)v2 v = 0.19 m/s

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**Spring Energy: Example 1b**

What is the potential energy and kinetic energy at x = 2.0 cm? PE = ½ kx2 PE = ½ (20 N/m)(0.020 m2) = 4 X 10-3 J ½ kA2 = ½ mv2 + ½ kx2 ½ mv2 = ½ kA2 - ½ kx2 KE = 9 X 10-3 J - 4 X 10-3 J = 5 X 10-3 J

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**Spring Energy: Example 1c**

At what position is the speed 0.10 m/s? (Ans: cm)

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**Spring Energy: Example 2a**

A spring stretches m when a kg mass is suspended from it (diagrams a and b). Find the spring constant. (Ans: N/m)

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**Spring Energy: Example 2b**

The spring is now stretched an additional m and allowed to oscillate (diagram c). What is the maximum velocity?

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**The maximum velocity occurs through the origin:**

½ kA2 = ½ mv2+ ½ kx2 ½ kA2 = ½ mv2 (x=0 at the origin) kA2 = mv2 v2 = kA2/m v = \/kA2/m = \/(19.6 N/m)(0.100m)2/0.300kg v = m/s

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**Spring Energy: Example 2c**

What is the velocity at x = m? ½ kA2 = ½ mv2+ ½ kx2 kA2 = mv2+ kx2 mv2 = kA2 - kx2 v2 = kA2 - kx2 m v2 = 19.6 N/m(0.100m2 – m2) = 0.49 m2/s2 0.300 kg v = m/s

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**Spring Energy: Example 2d**

What is the maximum acceleration? The force is a maximum at the amplitude F = ma and F = kx ma = kx a = kx/m = (19.6 N/m)(0.100 m)/(0.300 kg) a = 6.53 m/s2

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**Trigonometry and SHM Ball rotates on a table**

Looks like a spring from the side One rev(diameter) = 2pA T = 2p m k f = 1 T

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**Period depends only on mass and spring constant**

Amplitude does not affect period vo = 2pAf or vo = 2pA T vo is the initial (and maximum) velocity

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Period: Example 1 What is the period and frequency of a 1400 kg car whose shocks have a k of 6.5 X 104 N/m after it hits a bump? T = 2p m = 2p (1400 kg/6.5 X 104 N/m)1/2 k T = 0.92 s f = 1/T = 1/0.92 s = 1.09 Hz

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Period: Example 2a An insect (m=0.30 g) is caught in a spiderweb that vibrates at 15 Hz. What is the spring constant of the web? T = 1/f = 1/15 Hz = s T = 2p m k T2 = (2p)2m k = (2p)2m = (2p)2(3.0 X 10-4 kg) = 2.7 N/m T (0.0667)2

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Period: Example 2b What would be the frequency for a lighter insect, 0.10 g? Would it be higher or lower? T = 2p m k T = 2p (m/k)1/2 T = 2p (1.0 X 10-4 kg/2.7 N/m)1/2 = s f = 1/T = 1/0.038 s = 26 Hz

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**Cosines and Sines Imagine placing a pen on a vibrating mass**

Draws a cosine wave

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x = A cos2pt or x = A cos2pft T A = Amplitude t = time T = period f = frequency

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**x = A cos2pft v = -vosin2pft a = -aocos2pft**

Velocity is the derivative of position Acceleration is the derivative of velocity

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Cos: Example 1a A loudspeaker vibrates at 262 Hz (middle C). The amplitude of the cone of the speaker is 1.5 X 10-4 m. What is the equation to describe the position of the cone over time? x = A cos2pft x = (1.5 X 10-4 m) cos2p(262 s-1)t x = (1.5 X 10-4 m) cos(1650 s-1)t

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**Cos: Example 1b What is the position at t = 1.00 ms (1 X 10-3 s)**

x = A cos2pft x = (1.5 X 10-4 m) cos2p(262 s-1) (1 X 10-3 s) x = (1.5 X 10-4 m) cos(1.65 rad) = -1.2 X 10-5 m

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**Cos: Example 1c What is the maximum velocity and acceleration?**

vo = 2pAf vo = 2p(1.5 X 10-4 m)(262 s-1) = 0.25 m/s

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F = ma kx = ma a = kx/m But we don’t know k or m a = k x Solve for k/m m T = 2p m k T2 = (2p)2m k = (2p)2 = (2p)2f2 m T2

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a = k x m a = (2pf)2x = (2pf)2A a = [(2p)(262 Hz)]2(1.5 X 10-4 m) = 410 m/s2

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Cos: Example 2a Find the amplitude, frequency and period of motion for an object vibrating at the end of a spring that follows the equation: x = (0.25 m)cos p t 8.0

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x = A cos2pft x = (0.25 m)cos p t 8.0 Therefore A = 0.25 m 2pft = p t 2f = 1 f = 1/16 Hz T = 1/f = 16 s

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**Cos: Example 2b Find the position of the object after 2.0 seconds.**

x = (0.25 m)cos p t 8.0 x = (0.25 m)cos p 4.0 x = 0.18 m

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**The Pendulum Pendulums follow SHM only for small angles (<15o)**

The restoring force is at a maximum at the top of the swing. q Fr = restoring Force

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**Remember the circle (360o = 2p rad) q = x L**

Fr = mgsinq at small angles sinq = q Fr = mgq q L x q mg Fr

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Fr = mgq Fr = mgx (Look’s like Hook’s Law F = -kx) L k = mg T = 2p m k T = 2p mL mg

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T = 2p L g f = 1 = g T 2p L The Period and Frequency of a pendulum depends only on its length

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**Swings and the Pendulum**

To go fast, you need a high frequency Short length (tucking and extending your legs) f = g 2p L decrease the denominator

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Example 1: Pendulum What would be the period of a grandfather clock with a 1.0 m long pendulum? T = 2p L g Ans: 2.0 s

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Example 2: Pendulum Estimate the length of the pendulum of a grandfather clock that ticks once per second (T = 1.0 s). T = 2p L g Ans: m

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**Damped Harmonic Motion**

Most SHM systems slowly stop For car shocks, a fluid “dampens” the motion

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**Resonance: Forced Vibrations**

Can manually move a spring (sitting on a car and bouncing it) Natural or Resonant frequency (fo) When the driving frequency f = fo, maximum amplitude results Tacoma Narrows Bridge 1989 freeway collapse Shattering a glass by singing

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**Wave Medium Mechanical Waves Electromagnetic Waves Require a medium**

Water waves Sound waves Medium moves up and down but wave moves sideways Electromagnetic Waves Do not require a medium EM waves can travel through the vacuum of space

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**Parts of a wave Crest Trough Amplitude Wavelength**

Frequency (cycles/s or Hertz (Hz)) Velocity v = lf

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**Velocity of Waves in a String**

Depends on: Tension (FT) [tighter string, faster wave] Mass per unit length (m/L) [heavier string, more inertia] v = FT m/L

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Example 1: Strings A wave of wavelength 0.30 m is travelling down a 300 m long wire whose total mass is 15 kg. If the wire has a tension of 1000 N, what is the velocity and frequency? v = N = 140 m/s 15 kg/300 m v = lf f = v/l = 140 m/s/0.30 m = 470 Hz

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**Transverse and Longitudinal Waves**

Transverse Wave – Medium vibrates perpendicular to the direction of wave EM waves Water waves Guitar String Longitudinal Wave – Medium vibrates in the same direction as the wave Sound

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**Longitudinal Waves: Velocity**

Wave moving along a long solid rod Wire Train track vlong= E Elastic modulus r Wave moving through a liquid or gas vlong = B Bulk modulus

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**Ex. 1: Longitudinal Waves: Velocity**

How fast would the sound of a train travel down a steel track? How long would it take the sound to travel 1.0 km? vlong= E = (2.0 X 1011/7800 kg/m3)1/2 r vlong = 5100 m/s (much fast than in air) v = x/t t = x/v = 1000m/5100m/s = 0.20 s

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**Earthquakes Both Transverse and Longitudinal waves are produced**

S(Shear) –Transverse P(Pressure) – Longitudinal In a fluid, only p waves pass Center of earth is liquid iron

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**Energy Transported by Waves**

Intensity = Power transported across a unit area perpendicular to the wave’s direction I = Power = P Area pr2 Comparing two distances: I1r12 = I2r22

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Intensity: Example 1 The intensity of an earthquake wave is 1.0 X 106 W/m2 at a distance of 100 km from the source. What is the intensity 400 km from the source? I1r12 = I2r22 I2 = I1r12/r22 I2 = (1.0 X 106 W/m2)(100 km)2/(400 km)2 I2 = 6.2 X 104 W/m2

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**Reflection of a Wave Hard boundary inverts the wave**

Exerts an equal and opposite force Loose rope returns in same direction

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**Continue in same direction if using another rope boundary**

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**Constructive and Destructive Interference**

Destructive Constructive Interference Interference

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**Constructive and Destructive Interference : Phases**

Waves “in phase” “out of phase” in between

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**Resonance Standing Wave – a wave that doesn’t appear to move**

Node – Point of destructive interference Antinode – Point of constructive interference (think “Antinode,Amplitude) “Standing waves are produced only at the natural (resonant) frequencies.”

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**Resonance: Harmonics Fundamental Lowest possible frequency**

“first harmonic” L = ½ l First overtone (Second Harmonic) Second overtone (Third Harmonic)

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**Resonance: Equations L = nln n = 1, 2, 3….. 2 f = nv = nf1 2L v = lf**

v = FT m/L

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Example 1: Resonance A piano string is 1.10 m long and has a mass of 9.00 g. How much tension must the string be under to vibrate at 131 Hz (fund. freq.)? L = nln 2 l1 = 2L = m 1 v = lf = (2.20 m)(131 Hz) = 288 m/s

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v = FT m/L v2 = FT FT = v2m = (288 m/s)2(0.009 kg) = 676 N L (1.10 m)

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**What are the frequencies of the first four harmonics of this string?**

f1 = 131 Hz 1st Harmonic f2 = 262 Hz 2nd Harmonic 1st Overtone f3 = 393 Hz 3rd Harmonic 2nd Overtone f4 = 524 Hz 4th Harmonic 3rd Overtone

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**Hitting a Boundary Both reflection and refraction occur**

Angle of incidence = angle of reflection q1 q2 q1 = q2 Reflected wave air water Refracted wave

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**Refraction Velocity of a wave changes when crossing between substances**

Soldiers slow down marching into mud sin q1 = v1 sin q2 = v2

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Example 1: Refraction An earthquake p-wave crosses a rock boundary where its speed changes from 6.5 km/s to 8.0 km/s. If it strikes the boundary at 30o, what is the angle of refraction? sin q1 = v1 sin 30o = km/s sin q2 = v2 sin q2 8.0 km/s q2 = 38o

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Example 2: Refraction A sound wave travels through air at 343 m/s and strikes water at an angle of 50. If the refracted angle is 21.4o, what is the speed of sound in water? (Ans: m/s)

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Diffraction Note bending of wave into “shadow region”

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**Diffraction Bending of waves around an object**

Only waves diffract, not particles The smaller the obstacle, the more diffraction in the shadow region

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Vibration and Waves AP Physics Chapter 11.

Vibration and Waves AP Physics Chapter 11.

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