String Fixed at Both Ends
Trending Questions
A string is stretched between fixed points separated by 75 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string
- 5 Hz
- 10 Hz
- 2 Hz
- 4 Hz
- 5 Hz
- 10 Hz
- 2.5 Hz
- 7.5 Hz
- 2
- 4
- 8
- 1
- 38
- 23
- 89
- 94
- 4 m
- 2 m
- 8 m
- 3 m
- 1n=1n1+1n2+1n3
- √n=√n1+√n2+√n3
- n=n1+n2+n3
- 1√n=1√n1+1√n2+1√n3
- 3l4
- 4l5
- l5
- l4
- 25 kg
- 125 kg
- 5 kg
- 12.5 kg
- 51.4 Hz
- 17.2 Hz
- 8.6 Hz
- 21.4 Hz
What are normal modes of oscillation?
- v∝√x
- v∝x
- v∝1x
- v∝1x2
- length of the pipe is 1516 m
- length of the pipe is 916 m
- the maximum pressure at the open end is P0
- the minimum pressure at the open end is P0
- strain is maximum at nodes.
- strain is minimum at nodes.
- strain is maximum at antinodes.
- amplitude is zero at all points.
- (L2)
- (L4) and (3L4)
- L6, L2 and 5L6
- L8, 3L8, 5L8
- c=1πa
- c=πa
- b=ac
- b=1ac
- 300.47N
- 425.25 N
- 500 N
- 247.27 N
A sono-meter wire supports a 4 kg load and vibrates in fundamental node with a tuning fork of frequency 416 Hz. The length of the wire between the bridges is now doubled. In order to maintain fundamental node, the load should be changed to
1 kg
16 kg
2 kg
8 kg
Statement I: A speech signal of is used to modulate a carrier signal of . The bandwidth requirement for the signal is .
Statement II: The sideband frequencies are and . In the light of the above statements, choose the correct answer from the options given below:
Both statement I and statement II are false
Statement I is false but statement II is true
Statement I is true but statement II is false
Both statement I and statement II are true
Use value of √11250≈106
- 70.6 Hz
- 80.7 Hz
- 35.6 Hz
- 40.3 Hz
A 1cm long string vibrates with fundamental frequency of 256 Hz. If the length is reduced to 14cm keeping the tension unaltered, the new fundamental frequency will be (in Hz) (string is fixed at both ends)
- 64
- 1024
- 256
- 512
- 2f
- 2√2f
- f2√2
- f
- 144 cm
- 176 cm
- 152 cm
- 200 cm
- 120 Ω
- 90 Ω
- 60 Ω
- 30 Ω
In a resonance tube, using a tuning fork of frequency 325 Hz, two successive resonance lengths are observed as 25.4 cm and 77.4 cm respectively. The velocity of sound in air
338 ms-1
328 ms-1
330 ms-1
320 ms-1
- 1.42
- 1.77
- 1.82
- 1.21
A wire of length and mass per unit length is put under the tension of . Two consecutive frequencies that it resonates at are: and . Then in meter is
How is the frequency of a stretched string related to its length?
A tuning fork of frequency 480 Hz is used to vibrate a sono-meter wire having natural frequency 240 Hz. The wire will vibrate with a frequency of
240 Hz
480 Hz
720 Hz
will not vibrate