Janzen hui lo2 standing waves
- 1. StandingWaves Janzen Hui – LO2
- 2. StandingWaves A standing wave is described by the equation D(x,t) = 2A sin(kx) cos (ωt) D(x,t) = 2A sin(2πx/λ) cos (2πt/T) The amplitude A(x) depends on the position of the standing wave A(x) = 2A sin(kx) = 2A sin(2πx/λ) Combining the two equations creates: D(x,t)= A(x)cos(ωt)
- 3. Amplitude – Nodes and Antinodes Nodes: points where the amplitude is zero A(x) = 0 Consecutive nodes are half a wavelength apart Antinodes: point of maximum possible amplitude of 2A A(x)=2A Consecutive antinodes are half a wavelength apart Adjacent antinodes and nodes are a quarter of a wavelength apart
- 4. StandingWaves on Strings Two ends of a string, with length L, are fixed at both ends. The amplitude at each end must be zero (when x=0 and when x=L) A(x)= 2A sin(2πx/λ) = 2A sin(2πL/λ) = 0 Since a node repeats ever half a wavelength or π 2πL/λ = mπ Where m is a positive nonzero integer Rearranging the equation can help find frequency: f=v/λ = mv/2L= m/2L√(T/μ)
- 5. The lowest frequency(f1) produces the longest wavelength, where λ=2L and is the first harmonic. It has zero nodes. It is also called fundamental frequency or first harmonic Each consequent frequency is given by the equation fm=mf1 Where ‘m’ refers to the positive nonzero integers ‘m’ also indicates the subsequent harmonics or resonant frequencies. f2 =2f1 is the second harmonic.
- 6. Question Set 1 The D string on a guitar has a length of 0.70m and a linear mass density of 7.86 Χ 10-4kg/m.To maintain the D note on the guitar, the string is kept at a tension of 98N. A)What is the frequency when the wave length is half it’s original length B)What is the frequency if the tension is decreased by 10N.
- 7. Answer Set 1 A) Using the equation: f= m/2L√(T/μ).The length is half the original so L equals to 0.35m.TheTensions is given as 98N and the linear mass density is 7.86 Χ 10-4kg/m. The frequency becomes roughly 504Hz B)The tension is decreased by 10N to result in a new tension of 88N. Similar to the above question, the equation: f= m/2L√(T/μ) is used, but this time the tension is 88N rather than 98N. The frequency is roughly 239Hz.
- 8. Question Set 2 A standing wave has a wavelength of 25cm and a frequency of 9Hz. It has an antinode that reaches up to 4cm. A) Determine the standing wave equation of the form: D(x,t) = 2A sin(kx) cos (ωt) B) Determine the first position where the amplitude first reaches 1.6cm
- 9. Answer Set 2 – Pt.1 A)The ‘k’ and the ‘ω’ in the equation for D(x,t) can be found by: K= 2π/λ, ω=2π/T and whereT=1/f k= 2π/25 ≈.025 T= 1/9, ω=2π/T = 2π(9) ≈ 56.55 The antinode corresponds to the ‘2A’ of the equation D(x,t) and it equals to 4cm Plugging these values into the equation gives the answer: D(x,t) = 4 sin(.25x) cos (56.55t)
- 10. Answer Set 2 – Pt.2 B) Using the equation: A(x) = 2A sin(kx) = 2A sin(2πx/λ), and the results from above we can find the position (x) where the amplitude is first at 1.6cm. 1.6cm = 4sin(.25x) X≈ 1.64cm