Advertisements
Advertisements
Question
A tuning fork A, marked 512 Hz, produces 5 beats per second, where sounded with another unmarked tuning fork B. If B is loaded with wax the number of beats is again 5 per second. What is the frequency of the tuning fork B when not loaded?
Advertisements
Solution
Given that, vA = 512
On loading, frequencies of B decreased
So, in the first case v1 = 5
⇒ vB = vA ± 5
When B is loaded, the frequency of B becomes,
⇒ vB = 512 ± 5
We can write it as,
⇒ vB = 507 or vB = 517
On load in the frequency of B decreased by 507 to lower value of the number of beats will increase so vB ≠ 507
If vB = 517 then its frequency decreases by 10 Hz by then the number of beats will also be the same as 512 – 507 = 5.
Thus, the frequency of the tuning fork B when not loaded is 517 Hz.
APPEARS IN
RELATED QUESTIONS
In Doppler effect of light, the term “red shift” is used for ______.
(A) frequency increase
(B) frequency decrease
(C) wavelength decrease
(D) frequency and wavelength increase
The working of RADAR is based on ...............
a) resonance
b) speed of a star
c) Doppler effect
d) speed of rotation of sun
State any four applications of Doppler effect
What are beats?
When the observer moves towards the stationary source with velocity, 'V1', the apparent frequency of the emitted note is 'F1'. When the observer moves away from the source with velocity 'V1', the apparent frequency is 'F2'. If 'V' is the velocity of sound in air and `"F"_1/"F"_2 = 4` then `"V"/"V"_1 = ?`
If two tuning forks A and B are sounded together, they produce 4 beats per second. When A is slightly loaded with wax, they produce 2 beats when sounded again. The frequency of A is 256 Hz. The frequency of B will be ______
A tuning fork 'A' produces 5 beats per second with a tuning fork of frequency 480 Hz. When a little wax is stuck to a prong of fork A, the number of beats heard per second becomes 2. What is the frequency of tuning fork A before the wax is stuck to it?
Two waves of wavelength 2 m and 2.02 m, with the same speed superimpose to produce 2 beats per second. The speed of each wave is ____________.
Two tuning forks have frequencies 380 Hz and 384 Hz respectively. When they are sounded together, maximum sound is heard. How long will it take to hear the minimum sound?
A set of 56 tuning forks are arranged in series of increasing frequencies. If each fork gives 4 beats with preceding one and the frequency of the last is twice than than that of first, then frequency of the first fork is ____________.
A pipe closed at one end has a length of 1m and at its open end, 0.25 m long uniform string is vibrating in its third harmonic and resonates with a fundamental frequency of pipe. If the tension in the string is 100 N and the speed of sound is 340 `m/s`, then the mass of the string is near ______
Two identical wires are vibrating in unison. If the tension in one of the wires is increased by 2%, five beats are produced per second by the two vibrating wires. The initial frequency of each wire is ______
`(sqrt(1 . 0.2) ≑ 1 . 0.1)`
A note produces 4 beat/s with a tuning fork of frequency 510 Hz and 6 beat/s with a fork of frequency 512 Hz. The frequency of the note is ______
When two waves of almost equal frequencies n1 and n2 reach at a point simultaneously, what is the time interval between successive maxima?
The velocity of sound in a gas in which two wavelengths 4.08 m and 4.16 m produce 40 beats in 12 s, will be ______.
Two tuning forks of frequencies 320 Hz and 340 Hz are sounded together to produce a sound wave. The velocity of sound in air is 326.4 m/s. Calculate the difference in wavelengths of these waves.
One beat means that the intensity of sound should be ______.
