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Find Amplitude and Frequency

  1. Nov 9, 2008 #1
    * A 200g block attached to a horizontal spring is oscillating with an amplitude of 3 cm and a frequency of 2 Hz. Just as it passes through the equilibrium point a shalock so trp blow exerts a force impulse on the that the speed of the block is increased by .5m/s. What are the new a) frequency and b) amplitude?

    My Attempt:

    total energy of the mass moving with amplitude A and oscillating with frequency f will be
    E = mA^2ω^2 / 2

    where ω = 2πf
    f = frequancy of oscillation
    A = amplitude of oscillation .
    at the mean postion the total eneregy will be in kinetic energy form
    E = mv^2 / 2
    v =velocity of the mass at equalibrium
    if 0.5m/sec is added to velocity

    Now the new total eneryg
    E' = m*(v + 0.5)^2 / 2 .
    m = mass of the block
    E' = mA'^2 ω^2 / 2
    the frequancy remains constanat
    and the amplidue to of the new oscillation changes
    ω = 2E' / mA'^2


    but at the end I am left with 2 unknown and i am not to solve this


    please help its urgent
     
  2. jcsd
  3. Nov 9, 2008 #2

    alphysicist

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    Homework Helper

    Hi dsptl,

    That's right; so what is the speed of the mass at the equilibrium point before the force impulse add the 0.5m/s? Then you will be able to find the value of the energy after the force impulse. What do you get?
     
  4. Nov 9, 2008 #3
    total energy of the mass moving with amplitude A and oscillating with frequency f will be
    E = mA^2ω^2 / 2 = (.2)(.03)^2(4pi)^2 / 2 = .0142J

    where ω = 2πf
    f = frequancy of oscillation
    A = amplitude of oscillation .

    at the mean postion the total eneregy will be in kinetic energy form
    E = mv^2 / 2
    .0142 = .2 v^2 /2

    v = .377m/s

    v =velocity of the mass at equalibrium
    if 0.5m/sec is added to velocity = .377m/s + .5m/s = .877m/s


    Is this right till this point? after this point I have no idea what to do....

    Now the new total eneryg
    E' = m*(v + 0.5)^2 / 2 .
    m = mass of the block
    E' = mA'^2 ω^2 / 2
    the frequancy remains constanat
    and the amplidue to of the new oscillation changes



    .
     
  5. Nov 9, 2008 #4

    alphysicist

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    Perhaps I'm not understanding what you're asking, but you now can solve for A', since you have everything else.
     
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