Beat Frequency
Trending Questions
- 200
- 210
- 205
- 215
- 526 Hz
- 542 Hz
- 546 Hz
- 530 Hz
- 4
- 5
- 7
- 6
- 360
- 180
- 60
- 3
- 2
- 4
- 8
- 10
When the temperature of the air column decreases to 16∘C then one beat per second is produced with same tuning fork. The frequency of the tuning fork is
- 50 Hz
- 80 Hz
- 70 Hz
- 60 Hz
10 beats per second?
- 2.0 m/s
- 2.5 m/s
- 30 m/s
- 3.5 m/s
- 72 Hz
- 288 Hz
- 84 Hz
- 87 Hz
A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first one is now loaded with a little wax and the beat frequency is found to increase to 6 per second. What was the original frequency of the tuning fork?
A bus is moving with a velocity of 5 m/s towards a huge wall. the driver sounds a horn of frequency 165 Hz. If the speed of sound in air is 355 m/s, the number of beats heard per second by a passenger on the bus will be approximately,
6
5
4
3
- Open end will be an antinode
- Odd harmonics of the fundamental frequency will be generated
- All harmonics of the fundamental frequency will be generated
- Pressure change will be maximum at both ends
- 258 Hz
- 254 Hz
- 250 Hz
- 262 Hz
Define frequency
A pipe of length is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below . The velocity of sound in air is
- zero
- 6
- 12
- 3
- 1.2f and λ
- f and 1.2λ
- 0.8f and 0.8λ
- 1.2f and 1.2λ
- 196 Hz
- 200 Hz
- 209 Hz
- 202 Hz
- 1 sec
- 2 sec
- 1/4 sec
- 1/2 sec
An air column in a pipe, which is closed at one end, will be in resonance with a vibrating body of frequency 166 Hz, if the length of the air column is
0.50 m
2.00 m
1.50 m
1.00 m
- 420
- 460
- 360
- 320
- 268 Hz
- 250 Hz
- 260 Hz
- None of these
Reason: The principle of superposition is valid only if the frequencies of the oscillators are nearly equal.
- Assertion and reason both are correct.
- Assertion and reason both are wrong
- Assertion is correct while reason is wrong
- Assertion is wrong while reason is correct.
- 255 Hz
- 320 Hz
- 245 Hz
- 420 Hz
- 6 Hz
- 15 Hz
- 17 Hz
- 19 Hz
A musical instrument is made using four different metal strings. 1, 2, 3 and 4 with mass per unit length μ, 2μ, 3μ and 4μ respectively. The instrument is played by vibrating the strings by varying the free length in between the range L0 and 2L0 . It is found that in string- 1 ( μ) at free length L0 and tension T0 the fundamental mode frequency is f0. List-I gives the above four strings while List-II lists the magnitude of some quantity.
List-I | List-II | ||
(i) | String-1(μ) | (P) | 1 |
(ii) | String-2(2μ) | (Q) | 1/2 |
(iii) | String- 3(3μ) | (R) | 1/√2 |
(iv) | String-4(4μ) | (S) | 1/√3 |
(T) | 3/16 | ||
(U) | 1/16 |
If the tension in each string is T0, the correct match for the highest fundamental frequency in f0 units will be,
I→P, II→R, III→S, IV→Q
I→Q, II→P, III→R, IV→T
I→Q, II→S, III→R, IV→P
I→P, II→Q, III→T, IV→S
A and B are simultaneously excited, 5 beats per second are produced. Then the frequency of the tuning fork A (in Hz) is