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Homework Help: Harmonic motion unknowns

  1. Nov 10, 2005 #1
    Hi,

    I'm having trouble with this problem because I don't know how to deal with all these unknowns:

    4. A block attached to a spring with unknown spring constant oscillates with a period of 2.0 s. What is the period if :

    a. The mass is doubled?
    b. The mass is halved?
    c. The amplitude is doubled?
    d. The spring constant is doubled?

    I think I should use T = 2π * sqrt( m / k ), but what are m and k... ??

    Also, what relates this to amplitude?


    Hopefully I'm doing this problem correctly, but I don't know how to get acceleration:

    7. The bow of a destroyer undergoes simple harmonic vertical pitching motion with a period of 8.0 s and an amplitude of 2.0 m.

    a. What is the maximum vertical velocity of the destroyer’s bow?

    f = 1/T = .125rev/s

    ω = 2πf = .785rad/s

    v = ωr = (.785rad/s) (2m) = 1.57m/s

    b. What is the maximum acceleration?

    ??

    c. An 80 kg sailor is standing on a scale in the bunkroom in the bow. What are the maximum and minimum readings on the scale in newtons?

    80kg * ( 9.8m/s^2 + a)

    80kg * ( 9.8m/s^2 - a)


    Thanks,

    dusty.......
     
  2. jcsd
  3. Nov 10, 2005 #2
    m = the mass, k = spring constant

    You will have to write your answers in terms of 'T'.

    Well amplitude isn't in the equation for the period, so does it really affect T?

    I have to go to a lecture now... I'll finsh this reply when I return.
    (Quickly: if x = A Sin( wt),

    Then - dx/dt = v = Aw Cos(wt) - You noticed that v is a maximun at Cos(wt) = 1. (You used 'r' for amplitude, whereas I use 'A').

    So dv/dt = a. Hope this helps,

    Regards,
    Sam
     
  4. Nov 10, 2005 #3
    Ok,

    I'm not sure if I understood, but I've reworked the problems:

    For this problem, I hope I used logical mathematics and not contrived black magic...

    4. A block attached to a spring with unknown spring constant oscillates with a period of
    2.0 s. What is the period if :

    T = 2π * sqrt( m / k )

    a. The mass is doubled?

    T = 2π * sqrt( 2 * (m / k) )

    T = sqrt( 2 ) ( 2π * sqrt( m / k ) )

    T = sqrt( 2 ) * 2s

    T = 2.83s

    b. The mass is halved?

    T = sqrt( .5 ) * 2s

    T = 1.41s

    c. The amplitude is doubled?

    2s – not dependant on amplitude

    d. The spring constant is doubled?

    T = sqrt( .5 ) * 2s

    T = 1.41s


    So, do you mean that a = v[max] / period ?

    7. The bow of a destroyer undergoes simple harmonic vertical pitching motion with a period of 8.0 s and an amplitude of 2.0 m.

    a. What is the maximum vertical velocity of the destroyer’s bow?

    f = 1/T = .125rev/s

    ω = 2πf = .785rad/s

    v = ωr = (.785rad/s) (2m) = 1.57m/s

    b. What is the maximum acceleration?

    a = Δv / Δt

    a = (1.57m/s) / (8s) = .2m/s^2

    c. An 80 kg sailor is standing on a scale in the bunkroom in the bow. What are the
    maximum and minimum readings on the scale in newtons?

    F = ma


    F[max] = 80kg * ( 9.8m/s^2 + .2m/s^2 ) = 800N

    F[min] = 80kg * ( 9.8m/s^2 - .2m/s^2 ) = 768N


    And here is an aditional problem which I've worked, if you don't mind checking that I'm on the right path.

    6. The position of a particle is given by 0.07 cos (6п t) m, where t is in s.

    a. What are the frequency and the period?

    ω = 2πf = 6π

    f = ω / 2π = 6π / 2π

    f = 3rev/s


    T = 1/f = .33s

    b. What is the amplitude?

    .07m

    c. What is the maximum speed?

    v = rω = (.07m) (6π rad/s)

    v = .0037m/s

    d. What is the maximum acceleration?

    a = Δv / Δt

    a = (.0037m/s) / (.33s) = .0112m/s^2

    e. What is the first time after t = 0 that the particle is at the equilibrium position?

    cos (6π*t) = 0

    6π*t = cos^-1 ( 0 )

    t = cos^-1 ( 0 ) / 6π

    t = .083s

    f. What is the first time the particle is at x = 0 and moving to the right?

    cos (6π*t) = 0

    6π*t = cos^-1 ( 0 ) = (1/2)π, but going left

    (1/2)π + π = (3/2)π

    t = (3/2)π / 6π

    t = .25s

    Thanks for all the help,

    dusty.......
     
  5. Nov 10, 2005 #4
    I agree with all of your answers to question 4.

    I'm sorry, I seem to have confused you with Question 7, let me try again...

    We can say the the position x for an object undergoing SHM can be written as:

    x = A Sin (wt)

    By definition, if we differentiate a function of displacement with respect to time (dx/dt), we get velocity. So...

    v = dx/dt = d/dt(A Sin wt) = Aw Cos wt

    Also by definition if we diferentiate a function of velocity with respect to time (dv/dt), we get acceleration. So can you now differentiate:

    d/dt(Aw Cos (wt)) If you can't don't worry... I'm just trying to boost your knowledge of SHM. If you understand this, then great! If not, let me tell you that:

    amax = Aw2 (this equation is given in exams that I have taken in sixth-form and University).

    Let me know how you get on,
    Sam
     
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