**Unformatted text preview: **PHYS2405 Chapter 4: Resonance Solutions Spring 2019 Question 1 Three pure tones of frequencies 12 Hz, 3 Hz, and 2 Hz are played one at a time near a taut string
of length 1 ft that is tied at both ends. It is observed that only the 12 Hz tone makes the string vibrate in one of its
allowed modes. From this information, which of the following could we deduce?
a.
b.
c.
d. the LNRF of the string.
all the allowed modes of vibration of the string.
the value of the wave velocity for the string.
the values of all the allowed wavelengths. Solution
So 12 Hz corresponds to a resonant frequency of the string, but 3 Hz and 2 Hz do not. Since we don’t know if the
12 Hz corresponds to the LNRF, it could corresponds to an overtone of the string. This means the LNRF could be
4 Hz or 6 Hz, the only other factors of 12 besides 2 and 3. But knowing = 1 f t and the string is fixed at both ends,
then
2 2(12 Hz)(1 f t) 24 tt/s = 2 → = = = . Since we don’t know the shape of the string, we don’t know in which mode it vibrates and therefore do not have a
value for . Therefore, the only thing we can conclude are the allowed wavelengths, because for a string fixed at
each end: = 2/. Even without knowing in which mode it vibrates at 12 Hz, we can find all the wavelengths. Question 2 Three pure tones of frequencies 12 Hz, 7 Hz, and 11 Hz are played one at a time near a taut string
of length 1 ft that is tied at both ends. It is observed that only the 12 Hz tone makes the string vibrate in one of its
allowed modes. From this information, which of the following could we deduce?
a.
b.
c.
d. that the LNRF of the string is 12 Hz.
that 12 Hz is either the LNRF or one of the other natural resonant frequencies of the string.
the value of the wave velocity for the string.
the values of all the allowed wavelengths. Solution
All we know is that 12 Hz corresponds to an allowed mode, not necessarily the LNRF. Since answer (b) uses the
word either, (b) is not the best answer because it provides the possiblity that 12 Hz is the LNRF. We do not know
in which mode it vibrates, anbd therefore we do not conclusively know the wavelength: it is one possible value of = 2/. So we can find all the allowed wavelengths (since it is fixed at each end) but nothing more. To find the
wave speed, we need frequency and wavelength or additional information the question did not provide. Question 3 A piece of wire is cut into two pieces, A and B, which are then tightly stretched and mounted between
two rigid walls. A and B have the same stretched lengths, but A is stretched more tightly than B. An external driving
source that is able to excite the wires is placed nearby. Which of the following could be possible situations. State the
circumstances necessary for the item to be true.
a.
b.
c.
d. A and B have all of the same natural resonant frequencies.
A and B share no natural resonant frequencies.
The LNRF of A matches a higher resonant frequency of B.
The LNRF of B matches a higher resonant frequency of A. Stephens PHYS2405 Chapter 4 Solutions, page 2 of 4 Spring 2019 Solution
We know that = = but > because wire A’s tension is higher. Looking at our equations, (A) = Stephens 2 and (A) = 2 ⟶ = PHYS2405 Chapter 4 Solutions, page 3 of 4 Spring 2019 or (A)
2 = (B) ⇒ (B)
2 (A) = <1 ⟹ (B) < (A) . a. It is impossible for both A and B to have all the same harmonics because the LNRF of B is smaller than the
LBRF of A.
b. It is possible for A and B to share no harmonics if and are incommensurate — they are not related by
ratios of whole numbers. For example, they would share some harmonics if
whole number = .
possibly different whole number To see this, if = 2 then their harmonics would have the same frequency if = 2 . But if we had
the situation where = then they would share no harmonics. c. This is possible, as mentioned in the previous part. d. This is impossible, as mentioned above, because we know that > implies that the LNRF of A is greater
than the LNRF of B. So only an overtone of B could match the LNRF of A. Question 4 A piece of wire is cut into two pieces, A and B, which are then tightly stretched and mounted between
two rigid walls. A and B have the same tension, but A is longer than B. If wire A is plucked and wire B nearby begins
to vibrate, it could be because
a.
b.
c.
d. the fundamental of A is equal to one of the natural resonant frequencies of B.
the LNRF of B is equal to one of the overtones of A.
the normal mode frequencies of A and B are the same.
none of these can be possible. Solution
The problems specifies = = since they have the same tension and are constructed from the same piece
of wire, but > . From our equations, (A) = 2 and (A) = 2 ⟶ = or
(A) 2 (B) 2 = ⇒ (B) (A) = >1 ⟹ (B) > (A) . So the LNRF of B is greater than the LNRF of A, but we don’t know how much greater since we are not given
additional information about the length of the wires. But since B vibrates when near A, one the modes of A must
equal a mode of B. However the LNRF of A cannot equal an resonant frequency of B nor can A and B have the
same modes since (B) > (A) . Though the LNRF of B could equal an overtone of A. Question 5 A boy has a mass on the bottom end of a spring and, holding the top end of the spring, can shake that
end up and down at any frequency he chooses. If he stretches the spring from the bottom and lets go without any
shaking, it oscillates at 5 Hz. He shakes his spring end under the following circumstances: in air (a) at 2 Hz, (b) at
5 Hz, (c) at 8 Hz; and then with the mass submerged in a bucket of water at (d) 5 Hz. If the amplitude of motion of the
boy’s hand is always the same, which of these circumstances will result in the largest amplitude of the mass on the
end of the spring? Can you tell which situation will have the second largest amplitude? Stephens PHYS2405 Chapter 4 Solutions, page 4 of 4 Spring 2019 Solution
Since the natural frequency of the spring is 5 Hz (we know this from how it oscillates after he extends and releases
it), if he drives the spring at this frequency then the system will be in resonance. This is not the case if the mass is
submerged in water. In that case, there is damping on the spring and the natural frequency will not longer be 5 Hz.
So the circumstance with the greater amplitude is driving at 5 Hz in air. Presumably, the oscillations in water will
be sufficiently damped (the text doesn’t inform you that the resonant frequency also changes with the addition of
damping), so that its amplitude will be less than when driven in air. But there’s not good reason to choose the 2 Hz
over the 8 Hz. Question 6 A string of length 2 ft is tied at both ends. When tuning forks of various frequencies are struck near the
string, it is found that only the 20 Hz and the 50 Hz forks will start the string vibrating. A knowledgeable friend tells
you that the 50 Hz mode on the string is 3 modes higher than the 20 Hz mode. What is the LNRF of the string? What
is the wave velocity of any wave on the string?
Solution
We could use heuristic arguments or algebra for this. Let’s go heuristic first. Since 20 Hz and 50 Hz are each modes,
then they must be share a common divisor. The largest choice is 10 Hz. Since 50 − 20 = 30 = 3 × 10, then the
LNRF is 10 Hz. If you aren’t scared of algebra, let’s also try it. Let be the mode number of the 20 Hz resonance.
Then the 50 Hz resonance must have a node number of ( + 3). So we can write each of these frequencies in
terms of the fundamental and we can determine the mode number of the 20 Hz tone:
20 Hz 50 Hz
6
20 Hz = 1 and 50 Hz = ( + 3)1 → 1 =
=
→ 2( + 3) = 5 ⇒ = = 2. +3
So 20 Hz corresponds to the second mode. Then 20 Hz = 21 means that 1 = 10 Hz. Stephens 3 ...

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- Spring '19
- STEPHENS
- Wavelength, LNRF