1 Physics -- Waves in Pipes and Strings

Explain the formation of the overtones in open and closed pipes.Antinode is formed at the open end and node at closed end. From the figure, let L be the length of the tube and the wavelength of the wave produced. The frequencies higher than fundamental frequency can be produced by blowi...

Explain the formation of the overtones in open and closed pipes.

Antinode is formed at the open end and node at closed end. From the figure, let L be the length of the tube and the wavelength of the wave produced. The frequencies higher than fundamental frequency can be produced by blowing air strongly at open end. Such frequencies are called overtones.

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What is the difference between a tone and a note?

Tone can be defined as the sound producing single frequency and the sound of several mixed frequency is called note.

Note are always pleasent to hear. Tone can be called as the pitch of more accurate frequency.

What happens to the frequency of transverse vibration of stretched string if its tension is halved and the area of a cross-section of the string is doubled?

Thefrequencyof transverse vibration of astretched stringis calculated as

where;

T is the tension on the string
μ is the mass per unit length

When thetensionis halved and theareaof cross section isdoubled, we calculate the frequencyas;

Thus, when thetensionis halved and theareaof cross section isdoubled, thefrequencyincreases by a factor of1.414.

A vessel is placed below a water tap. We can estimate the height of the water level reached in the vessel from a distance simply by listening to the sound. How?

The frequency of sound produced in an air column is inversely proportional to the length of the air column. As the level of water in the vessel increases, the lengthof air column above it decreases. Hence, the frequency of sound produced goes on increasing.

In the resonance tube experiment. The first and the second resonance positions were observed at 17 cm and 52.6 cm respectively. The tuning fork used was the frequency of 512 Hz and the temperature ...

Suppose l₁= first resonance length and l₂ being second resonance length. Also, v0 and v27 being velocity at 0Cand27Crespectively.

First resonance length (l1) = 17 cm = 0.17 m
Second resonance length (l2) = 52.6 cm = 0.526 m
Frequency of the tuning fork (f) = 512 Hz

At 27C,

v₂₇ = 2f(l2-l1)

= 2 x 512(0.526-0.17)

= 364.5 m/s

We know vT,

Also, End correction (c) =

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