- #1
MrWarlock616
- 160
- 3
1. Homework Statement :
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. Homework 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##
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. Homework 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##
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