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Sound wave no idea how to do it

  1. Jun 9, 2010 #1
    1. The problem statement, all variables and given/known data

    (a) A thin, taut string tied at both ends and oscillating in its third harmonic has its shape
    described by the equation
    y(x,t) = 5.60cm (sin[(0.0340 rad/cm)x]sin[(50.0 rad/s)t]),
    where the origin is at the left end of the string, the x-axis is along the string and the
    y-axis is perpendicular to the string.
    (i) Draw a sketch that shows the standing wave pattern. (3 marks)
    (ii) Find the amplitude of the two travelling waves that make up this standing wave.
    (2 marks)
    (iii) What is the length of the string? (2 marks)
    (iv) Find the wavelength, frequency, period, and speed of the travelling waves. (4 marks)
    (v) Find the maximum transverse speed of a point on the string. (2 marks)
    (vi)What would be the equation y(x,t) for this string if it were vibrating in its eighth
    harmonic? (3 marks)

    2. Relevant equations



    3. The attempt at a solution

    the only thing i can think to do with this is stick the whole thing in my 89 and do it like that??
     
  2. jcsd
  3. Jun 9, 2010 #2
    I recognise the exam is tommorow so here is what I got from another source.


    Solution:
    Visualize: What does the third harmonic mode look like ?
    y(x,t) = (5.60cm) sin [(0.0340 rad/cm)x] sin [(50.0 rad/s)t]
    Compare with standing wave expression
    y = 2A sin kx sin ωt
    λ3
    L
    (a) Recall that a standing wave is the sum of two identical traveling waves moving opposite each other. At the anti-nodes, the two waves constructively interfere into a standing wave with an amplitude twice the original amplitude of either traveling wave (we showed this in class!!!) . In fact,
    y (for standing wave) = 2A sin kx sin ωt
    Thus, by comparison to the given equation, 2A = 5.6 cm so A = 2.8 cm
    (b) From the picture you should note that λ3 = 2L/3 or L = (3/2) λ3
    Or you can always get this from the expression for fundamental modes: f3 = 3(v/2L) = v/λ3.
    But the wavenumber k gives 2π/λ and examination of the expression above gives k = 0.034 rad/cm = 3.4/m so that λ = 2π/k = 2π/(3.4/m) = 1.85 m. Thus L = (3/2) λ3 = 2.78 m
    (c) The wavelength of traveling waves is the same as the standing wave: λ = 1.85 m
    (d) The frequency and period of the traveling waves is the same as the standing wave.
    Comparison of the sin ωt term with the given expression gives
    ω = 50 rad/s or f = ω/2π = 7.96 Hz.
    and T = 1/f = 0.1256 sec
    (e) The speed of the traveling wave is given by v = λf = (2.78 m)(7.96Hz) = 22.13 m/s
    (f) Maximum transverse speed of a point on the string can be found by taking dy/dt and maximixing this.
    dy/dt = (5.60 cm)(50 rad/s) sin [(0.0340rad/cm)x] [cos (50 rad/s)t]. The maximum values of sin and cos is 1 so the maximum dy/dt must be (5.6 cm)(50 rad/s) = 280 rad/
     
  4. Jun 9, 2010 #3
    cool thanks. how close to the actual exam do you reckon the practice test actually is??
     
  5. Jun 9, 2010 #4
    From what I have seen in the maths 11218 exam I would think very close
     
  6. Jun 9, 2010 #5
    ya that math exam was close to the practice and fricken easy
     
  7. Jun 9, 2010 #6
    hi in the solution you provided they have listed

    (e) The speed of the traveling wave is given by v = λf = (2.78 m)(7.96Hz) = 22.13 m/s

    is this correct? it seems to me the wavelength λ = 2pi/k or 2pi/3.4m = 1.85m thus v=1.85m * 7.96Hz = 14.70m/s

    comments?

    edit: also for question
    (vi)What would be the equation y(x,t) for this string if it were vibrating in its eighth
    harmonic? (3 marks)

    -I get amplitude remains the same
    k is = 2pi/lamda
    new k needs new lamda
    remember 8th harmonic, lamda = 2L/n with same string length
    lamda = 2*2.775m/8
    = 0.69375
    k= 2pi/0.69375
    = 9.05684rad/m
    /100
    0.0905rad/cm

    since we now have a new lamda we must have a new frequency and we know that wave speed is the same as it is a property of the string and so f=v/lamda f=14.7/0.69375 = 21.1892Hz
    so new frequency is 21.2Hz which is 21.2*2pi = 133.136rad/s

    so new formula is

    y(x,t) = (5.60CM) sin [(0.0905 rad/cm)x] sin [(133rad/s)t]

    sound about right?
    ps thanks for doing a fair bit of the question asking this semester pat666 ;)
     
    Last edited: Jun 9, 2010
  8. Jun 10, 2010 #7
    (e) The speed of the traveling wave is given by v = λf = (2.78 m)(7.96Hz) = 22.13 m/s

    is this correct? it seems to me the wavelength λ = 2pi/k or 2pi/3.4m = 1.85m thus v=1.85m * 7.96Hz = 14.70m/s

    comments?

    Thanks! you are right, didn't notice the mistake.
     
  9. Jun 10, 2010 #8
    i finally get how to do this f********* question, its not actually that hard once you see whats going on. thanks for all the help!!!
     
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