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Homework Help: Simple harmonic motion of a body

  1. Sep 23, 2004 #1
    My teacher gave us lots of exercise without answer and help i feel so frustrated to complete all those exercise but i can't answer any of them. i search in the book and i can't find anyway suitable way for that question. this question is very simple but i can't get a away to answer it. please help.

    1)A body describing simple harmonic motion has a maximum acceleration of 8pie m/s2 and a maximum speed of 1.6 m/s. find the period T and the amplitude R

    2)A size is swinging in a horizontal circle 0.8m in diameter, at 30rev/min. A distant light causes a shadow if the stone to be formed on a nearby wall. what is the amplitude of the motion of the shadow? what is the frequency? what is the period?

    3)a body executes SHM with an amplitude of 2.00cm and a frequency of 3.00Hz. At t=0s, the displacement is x(0)=0cm and the velocity v(0) is positive.
    (a) Obtain analytical expressions for the displacement x(t), the velocity v(t), the acceleration a(t)
    (b)evaluate the expressions found in part (a) for t=50ms

    anyone who is concern to help me can also add me at msn suding_junix@hotmail.com
    your help in solving this questions is highly appreciated thanks...
  2. jcsd
  3. Sep 23, 2004 #2
    1) Think about a spring with a mass at the end, which is compressed along a horizontal surface. It is fully compressed then released. Where is the acceleration at its max? Where is its speed at a max? Always write out a list of the information you have, then try to find equations that relate the variables concerned.

    2) Hint: the shadow will look and act like a spring.

    3) Try thinking about the sinusoidal graphs made by SHM. After that, try converting from one graph (x-t) to the next (v-t). How are amplitude and frequency represented graphically?
  4. Sep 23, 2004 #3
    Problem #1:

    [tex] x(t)=x_{max}\cos \left( \omega t + \phi \right) [/tex]
    [tex] \frac{dx}{dt}=v(t)=-\omega x_{max}\sin \left( \omega t + \phi \right) =-v_{max}\sin \left( \omega t + \phi \right) [/tex]
    [tex] \frac{dv}{dt}x(t)=-\omega ^2 x_{max}\cos \left( \omega t + \phi \right)=-a_{max}\cos \left( \omega t + \phi \right) [/tex]

    So, we have

    [tex] \frac{a_{max}}{v_{max}}=\frac{\omega ^2 x_{max}}{\omega x_{max}}=\omega [/tex]

    and also

    [tex] \omega = 2\pi f=\frac{2\pi}{T}\Longrightarrow T = \frac{\omega}{2\pi}= \frac{1}{2\pi}\cdot \frac{a_{max}}{v_{max}} = \frac{1}{2\pi}\cdot \frac{8\pi \frac{m}{s^2}}{1.6\frac{m}{s}}=2.5 s [/tex] (Period)

    [tex] v_{max}=\omega x_{max} \Longleftrightarrow x_{max}=\frac{v_{max}}{\omega}=\frac{1.6 \frac{m}{s}}{\frac{2\pi}{2.5 s}}=\frac{2}{\pi}m [/tex] (Amplitude)

    Problem #2

    (a) [tex]f=30min^{-1}[/tex]

    (b) [tex]T = \frac{1}{f}=\frac{1}{30}min [/tex]

    Problem #3

    We are given:

    [tex] x_{max}=2.00 cm [/tex]
    [tex] f=3.00 Hz [/tex]
    [tex] t=0 [/tex]
    [tex] \phi =0 [/tex]
    [tex] x(0)=0 [/tex]
    [tex] v(0)>0 [/tex]
    [tex] \omega = 2\pi f = 2\pi\left( 3.00 Hz \right) [/tex]

    So, we find

    [tex] x(t)=x_{max}\cos \left( \omega t + \phi \right) = \left( 2.00 cm \right)\cos \left( \omega t + \phi \right) [/tex]

    [tex] \frac{dx}{dt}=v(t)=-\omega x_{max}\sin \left( \omega t + \phi \right) =-\left[ 2\pi \left( 3.00 Hz \right) \right] \left( 2.00 cm \right) \sin \left( \omega t + \phi \right)[/tex]

    [tex] \frac{dv}{dt}x(t)=-\omega ^2 x_{max}\cos \left( \omega t + \phi \right)=-\left[ 2\pi \left( 3.00 Hz \right) \right]^2 \left( 2.00 cm \right) \cos \left( \omega t + \phi \right) [/tex]

    At [tex] t=0 [/tex]:

    [tex] x(0)=x_{max}= 2.00 cm [/tex]

    [tex] v(0)=0[/tex] (not > 0)

    [tex] a(0)=-\omega ^2 x_{max}=-\left[ 2\pi \left( 3.00 Hz \right) \right]^2 \left( 2.00 cm \right)=12\pi \frac{m}{s} [/tex]

    At [tex] t=50ms=50\times10^{-3}s [/tex]:

    [tex] x(50\times10^{-3}s )= \left( 2.00 cm \right)\cos \left[ 2\pi\left( 3.00 Hz \right) \left(50\times10^{-3}s\right) \right]=\frac{\sqrt{-2\left(\sqrt{5}-5\right)}}{2}cm [/tex]

    [tex] v(50\times10^{-3}s )=-\left[ 2\pi \left( 3.00 Hz \right) \right] \left( 2.00 cm \right) \sin \left[ 2\pi\left( 3.00 Hz \right)\left(50\times10^{-3}s\right) \right]=-3\left( \sqrt{5}+1 \right)\pi \frac{cm}{s}[/tex]

    [tex] a(50\times10^{-3}s )=)=-\left[ 2\pi \left( 3.00 Hz \right) \right]^2 \left( 2.00 cm \right) \cos \left[ 2\pi\left( 3.00 Hz \right)\left(50\times10^{-3}s\right) \right]=-18\sqrt{-2\left(\sqrt{5}-5 \right)}\pi ^2 \frac{cm}{s^2} [/tex]
  5. Sep 23, 2004 #4
    I was hoping imbecile might arrive at those himself, but ok. Looks like you like Latexing! :)
  6. Sep 24, 2004 #5
    :redface: i appreciate so so so much for the solutions... i didn't look the answer clearly just go through it cause i am tired now!!! I will sure look into it after i wake up. so tired and stress up. whole day rushing my school homework, i try to finish up those i manage to do. actually that questions above i think of it for an hour and i get no solutions for that questions. my school teacher teach us the chapter but didn't teach us how to answer questions. i had asked my classmate whether they done the questions anot. the top student in class tells me that she just complete partially and got lots of questions she not sure how to do, while the rest don't know how to do. imagine doing questions without answer??? :surprised
    thank you guys for your help THANKS~~~
  7. Sep 24, 2004 #6
    I love the use of the term, simple, i dont think any harmonic motion is simple
  8. Sep 25, 2004 #7
    Hi Tiago,
    You had stated that ----> T = \frac{\omega}{2\pi}
    I had check... is it supposed to be
    -----> T = \frac{2\pi}{\omega} ???
  9. Sep 25, 2004 #8


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    Homework Helper

    Period or T is [tex] \frac{2 \pi}{\omega} [/tex]

    Frequency is [tex] \frac{\omega}{2 \pi} [/tex]
  10. Sep 28, 2004 #9
    Sorry for the delay in replying to your message.

    Oh... you're right. Sorry about that. I guess I was distracted. :)

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