Physics 1251 The Science and Technology of Musical Sound Unit 1 Session 5 Simple Harmonic Oscillators Unit 1 Session 5 Simple Harmonic Oscillators

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Foolscap Quiz: What is the wavelength of a C 4 (262. Hz) at room temperature? Frequency = 262. Hz; velocity = 343. m/s v = f ⋅ λ= (262. Hz) ⋅ λ = 343. m/s λ = (343. m/s) / (262. Hz) = m Frequency = 262. Hz; velocity = 343. m/s v = f ⋅ λ= (262. Hz) ⋅ λ = 343. m/s λ = (343. m/s) / (262. Hz) = m Name, Date,Session 5 Seat # Joe College Session 5 1/26/2002 #309 λ= 1.31 m (3 significant figures)

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Sound is a pressure/displacement wave that propagates in a material medium. But what causes sound to begin with? What is the sound of one hand clapping?

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Oscillating Hand Demonstration ☞♫Compression Rarefaction

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators An oscillation of a body causes a sound.

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators 1′ Lecture: A Simple Harmonic Oscillator is a device that oscillates at one frequency, determined by the spring constant k and the mass m of the system. A Simple Harmonic Oscillator is a device that oscillates at one frequency, determined by the spring constant k and the mass m of the system. The natural frequency of an SHO is related to these quantities by the equation: The natural frequency of an SHO is related to these quantities by the equation: f = 1/(2π) ‧ √ (k/m)

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Existential Physics Questions: What happens when you stretch a “spring”? What happens when you stretch a “spring”? What happens if you stretch the “spring” twice as much? Or three times as much? What happens if you stretch the “spring” twice as much? Or three times as much? What is a “spring”? What is a “spring”? What has this got to with acoustics? What has this got to with acoustics?

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators F x 1 N 2 N 3 N 4 N -1 N -2 N -3 N -4 N -.3 m -.4 m -.2 m -.1 m.2 m.2 m.1 m.1 m.3 m.3 m.4 m.4 m

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators F x xFxF0000xFxF m-1 N.2 m-2 N.3 m-3 N F/x = - 10 N/m

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Hooke’s Law When you stretch or compress a “spring,” the force (F) produced is proportional to the displacement (x) and in the direction to restore the system (-) to the original position.

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Spring Constant k The restoring force produced by a spring is proportional to the negative of its extension (or compression). F = - k ‧ x (Hooke’s Law) F : restoring force k: spring constant x: extension, the “stretch or squeeze (if negative).” What is the restoring force of a spring with k =25. N/m when it is stretch by 10 cm? F = - (25. N/m )(0.10 m) = -2.5 N

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Mass on a Spring Force Spring ————————→ Mass ——————————→ F = - k ‧ x x - k ‧ x

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators F x F/x = - 10 N/m F/x = N/m = -k F/x = N/m

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators x What determines the frequency of the oscillation of a simple harmonic oscillator? The stronger the force (k), the more rapid is the oscillation. The greater the mass (m) the slower is the oscillation.

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators 80/20 Frequency of Oscillation of Mass on a Spring f = 1/(2π)√(k/m) f = ‧ √ (k/m)

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators If mass = kg and k = 25.0 N/m, what will be the frequency of oscillation? What will be the period? f = 1/(2π) ‧ √ (k/m) f = 1/(2π) ‧ √ (k/m) =1/(2π) √ [(25.0 N/m)/(0.500 kg)] = √ [50.0] = 1.13 Hz P = 1/ f = 1/ (1.13 Hz) = sec

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators If we increase the mass from 0.5 kg to 1.0 kg what will happen to the frequency? Will it increase, decrease or stay the same? If it changes, by how much will it change? f = 1/(2π) ‧ √ (k/m) f ∝ √ (1/m) → f 2 / f 1 = √(m 1 /m 2 ) = √(0.5/1.0) ) = √(0.50) =0.71

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators If we double the springs what will happen to the frequency? Will it increase, decrease or stay the same? If it changes, by how much will it change? f = 1/(2π)√(k/m) f ∝ √k → f 2 / f 1 = √(k 2 /k 1 ) = √(50./25.) ) = √(2.0) =1.41 = √(50./25.) ) = √(2.0) =1.41

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Simple Harmonic Oscillators Tuning fork Tuning fork

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Simple Harmonic Oscillators Tuning Fork Tuning Fork (Bulova Accutron) (Bulova Accutron) Tuning Fork

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Simple Harmonic Oscillators “Mouth Harp” Oscillator

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Simple Harmonic Oscillators Kalimba Kalimba (Finger Piano)

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Simple Harmonic Oscillators Harmonica Harmonica (“Mouth Organ”)

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Simple Harmonic Oscillators Helmholtz Resonator Helmholtz Resonator

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Hermann von Helmholtz © American Institute of Physics, used by permission ( ) Prominent 19 th century physicist and mathematician. Author of Perception of Tone, highly influential treatise in musical acoustics.

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators A Helmholtz Resonator is a simple harmonic oscillator that uses air in a narrow neck as a mass and air trapped in a volume as a spring. Air Mass Air Spring

Physics 1251 Unit 1 Session 5 Simple Harmonic oscillators Examples of Helmholtz Resonators: n Bottle n Acoustic Tile n Cinder Block n Ocarina

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Helmholtz Resonator Ocarina Ocarina

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Summary: A Simple Harmonic Oscillator is a device that oscillates at one frequency, determined by the spring constant k and the mass m of the system. A Simple Harmonic Oscillator is a device that oscillates at one frequency, determined by the spring constant k and the mass m of the system. The natural frequency of an SHO is related to these quantities by the equation: The natural frequency of an SHO is related to these quantities by the equation: f = 1/(2π) ‧ √ (k/m)

Physics 1251 Unit 1 Session 5 Simple Harmonic Oscillators Summary: A Helmholtz resonator is a Simple Harmonic Oscillator comprising a trapped volume of air that acts as a spring and a narrow neck that acts as a mass. A Helmholtz resonator is a Simple Harmonic Oscillator comprising a trapped volume of air that acts as a spring and a narrow neck that acts as a mass.