Natural frequency of tuning fork

  • Thread starter fruitl00p
  • Start date
  • #1
94
0

Homework Statement



Using a highly sensitive parabolic sound collector, Sara records the frequency of a tuning fork as it drops into the Grand Canyon. She drops the vibrating tuning fork form rest at t=0. She records a frequency of 1783.0Hz at t=8.590s. What is the natural frequency of the tuning fork? Use Vsound=343.0 m/s



Homework Equations



f'=f*[ v/(v-vs)]

The Attempt at a Solution



Ok, I know the that the frequency of listener is 1783.0 Hz, that the velocity of the listener is zero, and the velocity of sound is given. My biggest problem is figuring out the velocity of source.

I have tried
v=g*t= 9.8*8.590 = 84.182m/s,
but that turned out not to be the velocity. Then I thought I need to find the distance so I did
d=(1/2)*g*t^2 =(1/2)*9.8*8.590^2 = 361.56 m,
then used the d to find the v by simply doing
v=d/t = 361.56/8.590=42.091 m/s.
However that was incorrect. How do I solve for the velocity of the source?
 

Answers and Replies

  • #2
961
0
this is tricky and similar problem was recently discussed so a search may bear fruit. Its a subtle problem insofar as the distance from the listener and the speed of sound need to be accounted for. In other words, at t=8.59s, part of that t was needed to get back to her and so represents a velocity at an earlier time. That help?
 
  • #3
94
0
I found the previous thread dealing with this problem. At first I thought I understood but I am a bit puzzled. There was the suggestion that I need to figure out the speed of fork had when it emitted the sound that arrived at the detector at t= 8.590s.

The hint was: "If the fork falls for T seconds it travels a distance D. Sound emitted at that time takes D/c seconds to reach the detector. The total time must equal t=7.88 seconds."

I keep rereading this and I am not grasping the hint yet. So the fork falls for a certain amount of time and naturally travels a distance. So I find the distance the fork travels. Using that I find how long it takes to reach detector by using the distance found divided by the velocity of sound in order to find that time.

It takes x s to reach the detector, but the total time is in my case 8.59 s.
So far if I do
D= 1/2 *g*t = 1/2*o.e*8.59= 361.56 m
its takes D/c seconds or 361.56/343 = 1.054 s to reach detector.

Now I'm a bit stuck. There is a 1.054s delay? so though the total time is 8.59s, I subtract the 1.054 from 8.59s? Am I on the correct track?
 
  • #4
961
0
not quite (besides it's 1/2 gt^2) but getting there.

we don't know the exact delay as this depends on the distance fallen, but your eqn for delay, t'=y/343 is right. we also know y as function of t, thats
1/2at^2. we know that t+t1=8.59s
 
  • #5
94
0
I meant (1/2)*g*t^2=d. I got the answer 361.54m from that not the one I wrote incorrectly:blushing:

I'm a bit tired so I apologize if what I type down doesn't make sense.
What I am trying to find the moment when the detecor begins detecting fork. So finding the time delay helps me find the distance before the time delay began. I thought I could find the distance using the above equation, but if that is not correct, then I can't picture another way finding the "total distance".

So I find the time delay and use that and total time to find moment the detector begins detecting. From that I can find the distance at the moment. then do vs=sqrt2gd, or something like that. Is that still far off?
 
  • #6
961
0
FL, don't despair and given the range of questions over the w/e I gotta wonder what you been up to over the last few weeks.
Only envy on this end, but
equate
1/2gt^2 with (8.59-t)* 343. do you see what I have done?? we equate the same Y from fall and the time it takes to get back. Its a quadradic. Solve it and you're 80 percent done.
 
  • #7
30
0
FL, I'm just adding to what DD is driving at but sometimes it helps to hear it a little differently, so here goes: There are two events going on, the fork falling to a certain point, and the sound getting from that point back to Sara's ear, and what both events have in common is the distance that is covered. It does take different times for each event but those times are related (Down time+Up time= 8.59sec). So write out an equation for the unknown distance based on free fall (T=down time), and another equation based on the speed of sound and (T=up time). Now set these two expressions for distance equal to each other and you should have the quadratic equation that DD referred to. Just remember to write it all out in terms of just Tdown or Tup and you can use the quadratic formula to solve for time. Knowing Tdown, you can get your free fall velocity, just as you tried in the first place. Hope this helps.
 
  • #8
94
0
I had some homework problems that were due this evening so now I can come back to this hmwk problem! Ok, I did the quadratic formula
I ended up getting 4.9t^2 +343t -2946.37 =0 I ended up getting t =7.735s.

bdrosd, your explanation of T up and T down made me more confused unfortunately. I understand that the total time is 8.59 so the time its traveling down plus the time it takes for the detector to detect is equal to that total time. So what time did I end up finding? The time when the fork gets immediately detected?

Because after I found this time I applied it to the equation .5*9.8*t^2=d
then used that d to find vs. So I tried .5*9.8*7.735^2= 293.216 m
but then I used that d and found vs=sqrt2*g*d= sqrt 2*9.8*293.216=75.80 m/s

f'=[v/v+vs]*f
f'*[v+vs/v]=f
1783.0*[343+75.80/343]=f
f=2177 Hz

Did I finally figure it out?:confused:
 
  • #9
94
0
Actually bdrosd, I don't think it really confused me...I just needed to think about the scenario you illustrated. Well I hope I understood what you said!
 
  • #10
961
0
I didn't check the Doppler shift, but the time was right.
don't forget that v=at, so you could have saved yourself a step.
v=9.8*7.735=75.8
Good job!
 
  • #11
94
0
Thanks! I got the correct answer.
 

Related Threads on Natural frequency of tuning fork

  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
9
Views
34K
  • Last Post
Replies
6
Views
19K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
7K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
4
Views
5K
Replies
1
Views
5K
  • Last Post
Replies
4
Views
2K
Replies
1
Views
3K
Top