# Few Problems (Waves, Elasticity, SHM)

1. Apr 16, 2012

### MrWarlock616

1. The problem statement, all variables and given/known data:
Hello PF! Here are a few problems I'm stuck on.

1. When a tuning fork and a source of frequency 200 Hz are sounded together, 4 beats/sec are heard. When the tuning fork is loaded with some wax, again 4 beats/sec are heard. Then the original frequency of the fork was:
(200 Hz, 204 Hz, 196 Hz, 220 Hz)

2. The vibrations of four air columns are shown in the figure. The ratio of their frequencies, i.e, $N_A:N_B:N_C:N_D$ is:
(1:2:4:3,
1:2:3:4,
4:3:2:1,
10:5:4:3
)

3. Pendulums A and B have periodic times 4 s and 4.25 s. They are made to oscillate simultaneously. At time t=0, they are in the same phase. After how many complete oscillations of A, they will be again in the same phase?
(7, 14, 21, 28)

4. When a 4kg mass is hung vertically from a light spring that obeys Hooke's law, the spring stretches by 2 cm. The work that should be done by an external agent in stretching the spring by 5 cm will be ($g = 10 m/s^2$):
(1.5 J, 1.75 J, 2 J, 2.5 J)

2. Relevant equations and the attempt at a solution:

Okay the first one is a bit stupid, because if wax is applied to the fork, then how can the same beats/sec be heard???? My book explains it like this: "Frequency of the fork may be 204 Hz or 196 Hz. When it is loaded, its frequency decreases. If it is 196, then the no. of beats will increase. If it is 204 Hz, it is possible to get 4 beats/sec." Can somebody explain me this? If the frequency of the fork decreases after applying wax, then the new frequency is either less than 204, or less than 196. Either way, the number of beats should not be 4 again.

For the second one, I got answer as 4:3:2:1, but my book has given the answer as 1:2:3:4. I need to know who's right.
Since $N_A=\frac{v}{4L_A}$ and $L_A=\frac{\lambda}{4}$ we get $N_A=\frac{v}{\lambda}$. Going like this we get $N_B=\frac{v}{2\lambda} ; N_C=\frac{v}{3\lambda} ; N_D=\frac{v}{4\lambda}$. So it should be 4:3:2:1 right?

For the third question, I couldn't do more than recognize the data. $n_a=\frac{1}{4}$ and $n_b=\frac{1}{4+\frac{1}{4}}=\frac{4}{17}$. But I'm stuck after that because it seems like some data is missing. The solution goes, "When A makes n oscillations, B makes (n-1) oscillations".. ???? how?

For the last one, I solved it like this:
Work $W=\frac{1}{2}(load)(extension)=0.5(40)(0.05)=1 J$

Last edited: Apr 16, 2012
2. Apr 16, 2012

### PeterO

I addressed 1 above - now to look at the others

3. Apr 16, 2012

### PeterO

See above

4. Apr 16, 2012

### PeterO

Did you ever consider that 196 is actually less than 204 ?

5. Apr 16, 2012

### MrWarlock616

Well, when I first looked at the problem, I instinctively said 204 Hz and it turned out to be correct. But I'm not sure why, because the beats cannot be 4 again after applying wax, because the new one will be 204 minus (some-value)

And yes the diagrams B and D are incorrect, but that's only because the waves are not properly drawn and they are not touching the walls of the column, right?

6. Apr 16, 2012

### MrWarlock616

OH RIGHT!! Thank you so much!!! For some unknown reason, I was thinking the new frequency will be between 200 and 204, but not less than 200.

7. Apr 16, 2012

### PeterO

Lets scale this up to represent two long distance runners. one can complete a lap of a standard athletics track in 40 seconds, the other takes 42.5 seconds.

They begin running side by side [equivalent of being in phase].
When A completes his first lap, B still have 2.5 seconds to go before completing his first lap.
OR
By the time B completes his first lap, A will be 2.5 seconds into his second lap.
By the time B completes his second lap, A will be 5 seconds into his third lap.
By the time B completes his third lap, A will be 7.5 seconds into his 4th lap.

After sufficient laps, A will be 40 seconds in front - which actually means side by side with B, but 1 lap ahead. A will have completed n laps, while B has completed n-1 laps.

8. Apr 16, 2012

### PeterO

No: B and D represent patterns you get in a pipe open at both ends - but these pipes have one closed end.
A & C represent the fundamental and first overtone for a stopped pipe - which equals the 1st and 3rd harmonic (you only get odd harmonics in a stopped pipe)

9. Apr 16, 2012

### PeterO

One Difficulty/error here is that the 40N load will only extend the spring 2cm. The total force we are looking at extends the spring 5 cm.

10. Apr 16, 2012

### MrWarlock616

PeterO, thank you so much for your replies! I understood the oscillations problem now..I got it now thanks

HOW could I not see that the diagrams were wrong?!? Exactly! ARGH

I also understood the last one. We can find out K and use it in $\frac{1}{2} K x^2$, x being the displacement. THANKS THANKS!!!!