NEET Waves Previous Year Questions with Solutions

In an open pipe of length L, both ends must be displacement antinodes (A). Standing wave modes require L = p * (lambda / 2), where p is the harmonic number.

• • Fundamental (1st harmonic, p=1): A - N - A ==> 1 node.
• • 2nd harmonic (p=2): A - N - A - N - A ==> 2 nodes.
• • 5th harmonic (p=5): exactly 5 displacement nodes.

In a closed pipe of length L, the closed end is a displacement node (N) and the open end is an antinode (A). Only odd harmonics exist: f_k = (2k - 1) * (v / 4L).

• • 1st harmonic (k=1): N - A ==> 1 node.
• • 3rd harmonic (k=2): N - A - N - A ==> 2 nodes.
• • 5th harmonic (k=3): 3 nodes.
• • 7th harmonic (k=4): 4 nodes.
• • 9th harmonic (k=5): exactly 5 displacement nodes (m = 5).

Substitute n = 5 and m = 5 into the ratio:

• • Both configurations have exactly 5 displacement nodes.
• • The ratio is identically 1, corresponding to Option (3).

For an organ pipe of length L open at both ends, displacement antinodes form at both open ends. The fundamental mode corresponds to a half wavelength in the pipe:

• • v is the speed of sound in air.

When the pipe is immersed in water to half its length (L/2), the water surface provides a rigid closed boundary at the bottom, while the top remains open. The vibrating air column has length:

• • The system is now a closed organ pipe of effective length L / 2.

In a pipe closed at one end and open at the other, a node forms at the water surface and an antinode at the open end:

• • The new fundamental frequency is:
• • f' = \frac{v}{\lambda'} = \frac{v}{2L} = f.
• • Thus the fundamental frequency remains unchanged at f, matching Option (2).

An open pipe of length L has antinodes at both open ends and a displacement node at the center. The pipe length accommodates half a wavelength:

A closed pipe of length L has a displacement node at the closed end and an antinode at the open end. The pipe length accommodates one-quarter of a wavelength:

Divide the two fundamental frequencies for pipes of identical length:

The beat frequency produced by two interfering acoustic sound waves is the absolute difference between their individual frequencies: $$f_{\text{beat}} = |f_A - f_B| = 6\text{ Hz}$$ Given that string A has a known frequency $$f_A = 530\text{ Hz}$$, there are two possible mathematical values for the initial frequency of string B ($$f_B$$): - Case 1: $$f_B = f_A + 6 = 530 + 6 = 536\text{ Hz}$$ - Case 2: $$f_B = f_A - 6 = 530 - 6 = 524\text{ Hz}$$

The fundamental frequency of a transverse wave on a stretched string of length $$L$$ and linear mass density $$\mu$$ is given by: $$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \implies f \propto \sqrt{T}$$ When the tension in string B is slightly decreased ($$T_B \downarrow$$), its frequency must definitely decrease ($$f_B' < f_B$$).

If $$f_B = 536\text{ Hz}$$, a decrease in tension causes $$f_B$$ to decrease (for instance, to $$535\text{ Hz}$$). Then: $$f_{\text{beat}}' = |530 - 535| = 5\text{ Hz}$$ The beat frequency would DECREASE. This directly contradicts the observed problem statement that the beat frequency INCREASES to 7 Hz.

If $$f_B = 524\text{ Hz}$$, a decrease in tension causes $$f_B$$ to decrease further away from $$f_A$$ (to $$523\text{ Hz}$$). Then: $$f_{\text{beat}}' = |530 - 523| = 7\text{ Hz}$$ The beat frequency INCREASES to 7 Hz, which perfectly matches the given experimental observation.

Therefore, the original frequency of string B was unequivocally 524 Hz.