Advertisements
Advertisements
प्रश्न
A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be is resonance with the wire?
Advertisements
उत्तर
Wire of twice the length vibrates in its second harmonic. Thus, if the tuning fork resonates at L it will resonate at 2 L This can be explained below The sonometer frequency is given by
`v = n/(2L) sqrt(T/m)` .....(n = number of loops)
Now, as it vibrates with length L, we assume `v = v_1`
`n = n_1`
∴ `v_1 = n_1/(2L) sqrt(T/m)` ......(i)
When length is doubled, then `v_2 = n_2/(2 xx 2L) sqrt(T/m)` ......(ii)
Dividing equation (i) by equation (ii), we get
`v_1/v_2 = n_1/n_2 xx 2`
To keep the resonance `v_1/v_2 = 1 = n_1/n_2 xx 2`
⇒ `n_2 = 2n_1`
Hence, when the wire is doubled the number of loops also gets doubled to produce the resonance. That is it resonates in the second harmonic.
APPEARS IN
संबंधित प्रश्न
Explain why (or how) In a sound wave, a displacement node is a pressure antinode and vice versa,
A 2 m long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is 200 m s−1 and the amplitude is 0⋅5 cm. (a) Find the wavelength and the frequency. (b) Write the equation giving the displacement of different points as a function of time. Choose the X-axis along the string with the origin at one end and t = 0 at the instant when the point x = 50 cm has reached its maximum displacement.
The number of possible natural oscillations of the air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz? (v = 340 m/s)
The displacement of a string is given by y (x, t) = 0.06 sin (2πx/3) cos (120 πt) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10−2 kg.
- It represents a progressive wave of frequency 60 Hz.
- It represents a stationary wave of frequency 60 Hz.
- It is the result of superposition of two waves of wavelength 3 m, frequency 60 Hz each travelling with a speed of 180 m/s in opposite direction.
- Amplitude of this wave is constant.
Which of the following statements are true for a stationary wave?
- Every particle has a fixed amplitude which is different from the amplitude of its nearest particle.
- All the particles cross their mean position at the same time.
- All the particles are oscillating with same amplitude.
- There is no net transfer of energy across any plane.
- There are some particles which are always at rest.
The wave pattern on a stretched string is shown in figure. Interpret what kind of wave this is and find its wavelength.
The pattern of standing waves formed on a stretched string at two instants of time are shown in figure. The velocity of two waves superimposing to form stationary waves is 360 ms–1 and their frequencies are 256 Hz.
- Calculate the time at which the second curve is plotted.
- Mark nodes and antinodes on the curve.
- Calculate the distance between A′ and C′.
Two identical strings X and Z made of same material have tension Tx and Tz in them If their fundamental frequencies are 450 Hz and 300 Hz, respectively, then the ratio `"T"_x/"T"_"z"` is ______.
Shown in the figure is rigid and uniform one meter long rod AB held in horizontal position by two strings tied to its ends and attached to the ceiling. The rod is of mass 'm' and has another weight of mass 2m hung at a distance of 75 cm from A. The tension in the string at A is :
Two travelling waves produce a standing wave represented by the equation. y = 1.0 mm cos (1.57 cm-1) x sin (78.5 s-1)t. The node closest to the origin in the region x > 0 will be at x = ______ cm.
A wire of length 2L is made by joining two wires A and B of the same length but different radii r and 2r, and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire A is p and that in B is q then the ratio p : q is ______.
Two closed end pipes when sounded together produce 5 beat per second. If their length are in the ratio 100 : 101, then fundamental notes produced by them are ______.
A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be ______ cm.
(Take the speed of sound in air as 340 ms-1.)
A tuning fork of frequency 480 Hz is used in an experiment for measuring the speed of sound (ν) in the air by resonance tube method. Resonance is observed to occur at two successive lengths of the air column, l1 = 30 cm and l2 = 70 cm. Then, ν is equal to ______.
A pipe open at both ends has a fundamental frequency f in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to ______.
