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2 questions about sound - logic reasoning needed

  1. Apr 3, 2010 #1
    Hi all,

    I was puzzled by the answers offered by an assessment book on this 2 questions. Hope fellow forummers can help me out on this please.

    1. A thin steel card is made to vibrate by holding it against the teeth of a cog wheel. The wheel is rotated with increasing speed. The sound wave is displayed on a cathode-ray oscilloscope.

    The question being asked is basically: what features of the sound will change as the steel card is hit with increasing speed.

    I would think that the answer is pitch (higher) and loudness (higher). But the answer offered by the book is pitch only. I am puzzled why loudness of sound is not increased. When i hit something quicker, i would think that more force is being imparted into the object. Does not an increase in speed translate to a preceding increase in force needed to make the wheel turn faster?

    2. A string is suspended with a weight tied to its end. The string is then reduced to a quarter of its original length, while the attached weight is reduced to half.
    The question asked whether the new setup will have a higher or lower pitch sound. The answer is higher pitch. My question is, does the weight attached and the length of the string follow a direct proportion relationship? In other words, if i half the length of the string, and half the attached weight, does the pitch of the sound stay the same?

    Hope forummers can help out this troubled physics teacher. :blushing:
     
  2. jcsd
  3. Apr 3, 2010 #2
    Regarding Q1.
    The loudness of the sound will depend on the amplitude of the waves. Increasing the speed of the cogwheel will increase the number of times the card vibrates per second, but will not increase its amplitude of vibration. I would imagine the amplitude would be increased by using a cog with larger teeth.

    Q2.
    The equation that governs the frequency of a stretched string is
    f = (1/2L)√(T/μ) where L is its length and T is the tension in it. μ is its mass per unit length and doesn't change in this case.
    So reducing the length to a half will increase the frequency by 2
    Reducing the tension by half (half the mass) will only reduce the frequency by √2
     
  4. Apr 3, 2010 #3
    Hi Stonebridge,

    Thanks for your wonderful insights to Q2.

    But i am still not fully convinced for Q1 though. My understanding is that amplitude is related to the energy contained within the waves. I see your point when you say that amplitude is increased through using a larger teeth wheel. But mass aside, wont there be more energy imparted to the steel card when hit by a "faster particle" as compared to a "slower particle"?
     
  5. Apr 3, 2010 #4
    There may be more energy imparted to the card; but does that mean the sound waves will be any louder? These are produced when the air near the card is moved, and if the card's amplitude doesn't change, the amplitude of the air vibrations doesn't change. The oscilloscope will show this amplitude.
    The sensation of loudness is something different. Our ears have different sensitivity to different frequencies, so the perceived loudness is somewhat subjective.
    What the oscilloscope will show is that the amplitude of the sound waves hitting the microphone hasn't changed.
    This does depend a little on the exact way the cogs hit the card, but I think the spirit of the question is that the answer required is that it's only the pitch that changes. Loudness is a tricky thing to quantify with sound, but the amplitude of the sound wave is simple to measure.
     
  6. Apr 3, 2010 #5
    The energy to deflect the card a certain distance depends solely upon the card and not at all upon the deflecting force.

    Consider what is suggested. The tooth of the cog impacts upon the edge of card, taking it a little way along its journey with itself. i.e. it deflects the card.
    At some point the card slips past the tooth and springs back to be caught by the next tooth.

    It does not matter how fast this happens the card is only deflected the same amount since it will always lose contact at the same point.

    BTW the energy in a wave is determined by the frequency and the amplitude.
     
  7. Apr 3, 2010 #6
    Thanks Stonebridge and Studiot. I am starting to see the light.

    Thank you for your responses.

    Both of you will make better teachers than me. :uhh:
     
  8. Apr 4, 2010 #7
    I believe the sound could definitely get louder. Just because the answer wasn't in the assessment doesn't mean you are wrong. It could just be the writers didn't think it was important, or something else.

    Think of a situation with a hammer and a piece of sheet metal. When you strike the metal harder, it makes a louder noise. I think the exact mechanism of this acoustic generation is somewhat complex, but the situation is essentially the same. You are correct to identify that a faster moving wheel imparts a larger impulse, in the same way as swinging the hammer harder. The larger the impulsive force, the more energy there is available to be dissipated through sound.

    One could think of a different situation where this might not be true, as with a hammer and a piece of rubber. The rubber might just absorb the hammer blow and dissipate the energy elastically without making a lot of noise, even if you swing the hammer hard. With the card example, however, it's somewhere in between a piece of metal and a piece of rubber.
     
    Last edited: Apr 4, 2010
  9. Apr 4, 2010 #8
    No, the energy is dependent on the strength of the impulsive load. When the wheel moves faster, this strength increases. Also, the card is not deflected the same amount every time anyways. When the wheel travels faster, it hits the card at faster intervals, and in general, at different states of card displacement.
     
  10. Apr 5, 2010 #9
    No the situation is not the same. It is essentially different.

    Your hammer remains in contact with your piece of sheet metal. Further the hammer contacts your sheet at essentially right angles, the ideal angle for maximum energy transfer.

    The cog tooth only deflects the card sufficiently to push past the it. What you are suggesting is the same as saying the cog's impact, which is remember is a glancing blow, is sufficient to not only push the card aside but also to push it clear of the cog.

    As soon as it is clear of the cog, there can be no further energy transfer.

    The situation at the contact point is also more complex because there are both tangential and normal forces acting at the point of contact between the tooth and the card. If you draw some (vector) diagrams you will see that the majority of the acceleration vector of the tooth ( and it is this vector which generates the force ) is tangential to the card.
     
  11. Apr 5, 2010 #10

    Pythagorean

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    With the same weight, something going faster does have more momentum and thus delivers a higher forced impulse if it comes to a complete stop.

    In the case of the cogwheel, which doesn't come to a complete stop, we have to assume a portion of the momentum is absorbed by the plate and the cogwheel is supplied with energy to restore it's momentum (almost immediately) and keep its speed constant.

    Now, if it's only partial momentum being absorbed, but the cogwheel is spinning faster, then it will make contact for a shorter time:

    F = dp/dt = m(dv/dt)

    either if dv (the momentum absorbed) gets bigger or if dt gets smaller, the resulting force is larger.

    I conclude from this that the amplitude will be increased, but I'm not completely sure of my method, as sheer forces may not work out as nicely as point particle collisions I've been taught with.
     
  12. Apr 5, 2010 #11
    Speed is not the issue.

    It takes a certain force, let us call this F, to deflect the card sufficiently to allow the tooth past.

    It does not matter how that force is applied.

    If the card is shifted to its deflected position more quickly it does not make any difference to this position it is still the same, consequently, F is still the same.

    Once the card has reached the slip past position, all the momentum of the tooth is directed parallel to the card and cannot exert further force. Like I said, draw the vector diagrams.

    I do not actually regard this as an impulsive force question. The length of time F is applied by the tooth is calculable and considerably longer than the time period over which impulsive foces are considered to act.
     
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