Standing Waves in Strings and Organ Pipes, Fundamental Mode and Harmonics, Beats NEET Questions

Standing Waves in Strings and Organ Pipes, Fundamental Mode and Harmonics, Beats

Standing Waves in Strings and Organ Pipes, Fundamental Mode and Harmonics, Beats MCQ Questions

7.
In a standing wave, particles on OPPOSITE sides of a node are:
A.
Phase difference of π/2
B.
Exactly in phase
C.
No fixed phase relationship
D.
Exactly out of phase — they always move in opposite directions (phase difference = Ļ€)
ANSWER :
D. Exactly out of phase — they always move in opposite directions (phase difference = Ļ€)
8.
The energy in a standing wave compared to two constituent progressive waves is:
A.
Half — destructive interference destroys half the energy
B.
Equal — total energy is conserved; energy is redistributed (concentrated near antinodes, zero near nodes) but total is preserved
C.
Zero — standing waves carry no energy
D.
Double — energy is created by constructive interference
ANSWER :
B. Equal — total energy is conserved; energy is redistributed (concentrated near antinodes, zero near nodes) but total is preserved
9.
A string of length L is fixed at both ends. The boundary conditions require:
A.
No restriction on wavelength — any wavelength is possible
B.
A node at one end and antinode at the other
C.
Nodes at both fixed ends (x = 0 and x = L), so L must equal an integer number of half-wavelengths: L = nĪ»/2
D.
Antinodes at both fixed ends
ANSWER :
C. Nodes at both fixed ends (x = 0 and x = L), so L must equal an integer number of half-wavelengths: L = nĪ»/2
10.
The allowed wavelengths for standing waves in a string of length L fixed at both ends are:
A.
Any wavelength Ī» such that Ī» < 2L
B.
Ī»_n = 2L/n for n = 1, 2, 3,… (wavelength must fit an integer number of half-wavelengths in L)
C.
Ī»_n = 4L/n for n = 1, 2, 3,…
D.
Ī»_n = 4L/(2nāˆ’1) for n = 1, 2, 3,…
ANSWER :
B. Ī»_n = 2L/n for n = 1, 2, 3,… (wavelength must fit an integer number of half-wavelengths in L)
11.
The allowed frequencies (normal modes) for a string of length L fixed at both ends with wave speed v are:
A.
ν_n = nv/(4L) for n = 1, 2, 3,…
B.
ν_n = (2nāˆ’1)v/(4L) for n = 1, 2, 3,… — only ODD harmonics
C.
ν_n = nv/(2L) for n = 1, 2, 3,… — ALL harmonics (both odd and even)
D.
Only one frequency ν₁ = v/(2L) is possible
ANSWER :
C. ν_n = nv/(2L) for n = 1, 2, 3,… — ALL harmonics (both odd and even)
12.
The FUNDAMENTAL MODE (first harmonic) of a string fixed at both ends has:
A.
Nodes at ends and centre; wavelength λ₁ = 2L/3
B.
One antinode at each end and node at centre; wavelength λ₁ = L
C.
One antinode at the centre; two nodes at the ends; wavelength λ₁ = 2L; frequency ν₁ = v/(2L)
D.
Two antinodes and one node at the centre; wavelength λ₁ = L
ANSWER :
C. One antinode at the centre; two nodes at the ends; wavelength λ₁ = 2L; frequency ν₁ = v/(2L)