101. In the expression \(\omega=2\pi f\), the quantity \(\omega\) is called
ⓐ. wavelength
ⓑ. angular frequency
ⓒ. wave number
ⓓ. linear mass density
Correct Answer: angular frequency
Explanation: Angular frequency \(\omega\) gives the rate of change of phase with time. It is related to ordinary frequency \(f\) by \(\omega=2\pi f\). The factor \(2\pi\) appears because one complete oscillation corresponds to \(2\pi\,\text{rad}\) of phase. Wavelength \(\lambda\) is a spatial repeat distance, while wave number \(k\) is related to wavelength by \(k=\frac{2\pi}{\lambda}\). Linear mass density \(\mu\) belongs to waves on strings and is not a time-rate of phase. The unit of \(\omega\) is commonly written as \(\text{rad s}^{-1}\).
102. A periodic wave has frequency \(50\,\text{Hz}\). Its angular frequency is
ⓐ. \(25\pi\,\text{rad s}^{-1}\)
ⓑ. \(50\pi\,\text{rad s}^{-1}\)
ⓒ. \(100\pi\,\text{rad s}^{-1}\)
ⓓ. \(2500\pi\,\text{rad s}^{-1}\)
Correct Answer: \(100\pi\,\text{rad s}^{-1}\)
Explanation: \( \textbf{Given data:} \) Frequency \(f=50\,\text{Hz}=50\,\text{s}^{-1}\).
\( \textbf{Required quantity:} \) Angular frequency \(\omega\).
\( \textbf{Relation:} \)
\[
\omega=2\pi f
\]
This relation converts cycles per second into radians of phase per second.
\( \textbf{Substitution:} \)
\[
\omega=2\pi(50\,\text{s}^{-1})
\]
\( \textbf{Calculation:} \)
\[
\omega=100\pi\,\text{s}^{-1}
\]
\( \textbf{Unit writing:} \)
\[
\omega=100\pi\,\text{rad s}^{-1}
\]
\( \textbf{Final answer:} \) The angular frequency is \(100\pi\,\text{rad s}^{-1}\).
Forgetting the factor \(2\pi\) would confuse ordinary frequency with angular frequency.
103. The wave number \(k\) of a sinusoidal wave is related to wavelength by
ⓐ. \(k=2\pi\lambda\)
ⓑ. \(k=\frac{2\pi}{\lambda}\)
ⓒ. \(k=\frac{\lambda}{2\pi}\)
ⓓ. \(k=f\lambda\)
Correct Answer: \(k=\frac{2\pi}{\lambda}\)
Explanation: Wave number \(k\) measures how rapidly the phase changes with position. One full wavelength \(\lambda\) corresponds to a phase change of \(2\pi\,\text{rad}\). Therefore, phase change per unit distance is \(k=\frac{2\pi}{\lambda}\). The unit of \(k\) is commonly \(\text{rad m}^{-1}\), or simply \(\text{m}^{-1}\) when radian is treated as dimensionless. The expression \(f\lambda\) gives wave speed \(v\), not wave number. A larger wavelength means a smaller value of \(k\), because the phase changes more slowly with distance.
104. A sinusoidal wave has wavelength \(0.40\,\text{m}\). The wave number is
ⓐ. \(0.20\pi\,\text{m}^{-1}\)
ⓑ. \(0.80\pi\,\text{m}^{-1}\)
ⓒ. \(2.5\pi\,\text{m}^{-1}\)
ⓓ. \(5\pi\,\text{m}^{-1}\)
Correct Answer: \(5\pi\,\text{m}^{-1}\)
Explanation: \( \textbf{Given data:} \) Wavelength \(\lambda=0.40\,\text{m}\).
\( \textbf{Required quantity:} \) Wave number \(k\).
\( \textbf{Relation used:} \)
\[
k=\frac{2\pi}{\lambda}
\]
This relation gives phase change per unit distance.
\( \textbf{Substitution:} \)
\[
k=\frac{2\pi}{0.40\,\text{m}}
\]
\( \textbf{Decimal handling:} \)
\[
0.40=\frac{2}{5}
\]
\( \textbf{Calculation:} \)
\[
k=2\pi\times\frac{5}{2}=5\pi\,\text{m}^{-1}
\]
\( \textbf{Final answer:} \) The wave number is \(5\pi\,\text{m}^{-1}\).
The reciprocal dependence means a shorter wavelength gives a larger wave number.
105. In a progressive sinusoidal wave, the phase may contain the term \(kx-\omega t+\phi\). The role of \(\phi\) is to represent
ⓐ. phase constant or initial phase
ⓑ. linear mass density of the string
ⓒ. maximum particle speed
ⓓ. pressure of the medium
Correct Answer: phase constant or initial phase
Explanation: The phase of a sinusoidal wave decides the stage of oscillation at a given position and time. In \(kx-\omega t+\phi\), the terms \(kx\) and \(\omega t\) describe how phase changes with position and time. The term \(\phi\) is a constant added to set the starting phase of the wave. It can shift the wave pattern without changing the amplitude, frequency, or wavelength. Linear mass density \(\mu\), particle speed, and pressure are different physical quantities. The phase constant helps locate the wave in its cycle at the chosen origin of \(x\) and \(t\).
106. Study the table and identify the mismatched quantity-unit pair.
| Row | Quantity | Usual unit |
|---|
| P | Angular frequency \(\omega\) | \(\text{rad s}^{-1}\) |
| Q | Wave number \(k\) | \(\text{m}^{-1}\) |
| R | Frequency \(f\) | \(\text{Hz}\) |
| S | Wavelength \(\lambda\) | \(\text{s}^{-1}\) |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: Wavelength \(\lambda\) is a distance, so its unit is \(\text{m}\), not \(\text{s}^{-1}\). Row P is suitable because angular frequency \(\omega\) is measured in \(\text{rad s}^{-1}\). Row Q is suitable because wave number \(k=\frac{2\pi}{\lambda}\) has unit \(\text{m}^{-1}\). Row R is suitable because frequency \(f\) is measured in \(\text{Hz}\), which is \(\text{s}^{-1}\). The mismatched row confuses a spatial period with a rate in time. Keeping \(\lambda\), \(f\), \(\omega\), and \(k\) separated is essential before using a progressive wave equation.
107. In the phase expression \(kx-\omega t+\phi\), the term \(kx\) is dimensionless because
ⓐ. \(k\) has unit \(\text{m}\) and \(x\) has unit \(\text{m}^{-1}\)
ⓑ. \(k\) has unit \(\text{m}^{-1}\) and \(x\) has unit \(\text{m}\)
ⓒ. \(k\) has unit \(\text{s}^{-1}\) and \(x\) has unit \(\text{s}\)
ⓓ. \(k\) has unit \(\text{kg m}^{-1}\) and \(x\) has unit \(\text{kg}\)
Correct Answer: \(k\) has unit \(\text{m}^{-1}\) and \(x\) has unit \(\text{m}\)
Explanation: The argument of a sine or cosine function must be dimensionless. In the phase term \(kx\), the coordinate \(x\) is a length and has unit \(\text{m}\). Therefore \(k\) must have the reciprocal length unit \(\text{m}^{-1}\), often written as \(\text{rad m}^{-1}\). Multiplying \(k\) and \(x\) gives a pure phase angle in radians. Similarly, \(\omega t\) is dimensionless because \(\omega\) has unit \(\text{rad s}^{-1}\) and \(t\) has unit \(\text{s}\). Dimensional consistency is a reliable way to separate wave number from angular frequency.
108. A sinusoidal wave has time period \(0.020\,\text{s}\) and wavelength \(0.50\,\text{m}\). The values of \(\omega\) and \(k\) are respectively
ⓐ. \(50\pi\,\text{rad s}^{-1}\) and \(2\pi\,\text{m}^{-1}\)
ⓑ. \(25\pi\,\text{rad s}^{-1}\) and \(4\pi\,\text{m}^{-1}\)
ⓒ. \(100\pi\,\text{rad s}^{-1}\) and \(0.25\pi\,\text{m}^{-1}\)
ⓓ. \(100\pi\,\text{rad s}^{-1}\) and \(4\pi\,\text{m}^{-1}\)
Correct Answer: \(100\pi\,\text{rad s}^{-1}\) and \(4\pi\,\text{m}^{-1}\)
Explanation: \( \textbf{Given data:} \) Time period \(T=0.020\,\text{s}\), wavelength \(\lambda=0.50\,\text{m}\).
\( \textbf{Angular frequency relation:} \)
\[
\omega=\frac{2\pi}{T}
\]
\( \textbf{Substitution for \(\omega\):} \)
\[
\omega=\frac{2\pi}{0.020\,\text{s}}
\]
\( \textbf{Calculation for \(\omega\):} \)
\[
\omega=100\pi\,\text{rad s}^{-1}
\]
\( \textbf{Wave number relation:} \)
\[
k=\frac{2\pi}{\lambda}
\]
\( \textbf{Substitution for \(k\):} \)
\[
k=\frac{2\pi}{0.50\,\text{m}}
\]
\( \textbf{Calculation for \(k\):} \)
\[
k=4\pi\,\text{m}^{-1}
\]
\( \textbf{Final answer:} \) \(\omega=100\pi\,\text{rad s}^{-1}\) and \(k=4\pi\,\text{m}^{-1}\).
The time period controls \(\omega\), while the wavelength controls \(k\).
109. A phase constant \(\phi\) is changed in the equation of a sinusoidal wave, while \(A\), \(k\), and \(\omega\) remain the same. The change mainly
ⓐ. changes only the initial phase of the wave
ⓑ. changes the wave speed by changing the medium
ⓒ. changes the wavelength without changing \(k\)
ⓓ. changes the frequency without changing \(\omega\)
Correct Answer: changes only the initial phase of the wave
Explanation: The phase constant \(\phi\) fixes the phase of the wave at the chosen origin of position and time. If \(A\), \(k\), and \(\omega\) remain unchanged, the amplitude, wavelength, and frequency remain unchanged. Changing \(\phi\) shifts the sinusoidal pattern in its cycle. It does not represent a change of medium or a change of propagation speed by itself. Since \(k=\frac{2\pi}{\lambda}\), unchanged \(k\) means unchanged \(\lambda\). Since \(\omega=2\pi f\), unchanged \(\omega\) means unchanged \(f\).
110. The phase of a sinusoidal wave at a point changes by \(6\pi\,\text{rad}\) in \(0.30\,\text{s}\). The angular frequency is
ⓐ. \(10\pi\,\text{rad s}^{-1}\)
ⓑ. \(18\pi\,\text{rad s}^{-1}\)
ⓒ. \(20\pi\,\text{rad s}^{-1}\)
ⓓ. \(6.3\pi\,\text{rad s}^{-1}\)
Correct Answer: \(20\pi\,\text{rad s}^{-1}\)
Explanation: \( \textbf{Given phase change:} \) \(\Delta \theta=6\pi\,\text{rad}\).
\( \textbf{Time interval:} \) \(\Delta t=0.30\,\text{s}\).
\( \textbf{Required quantity:} \) Angular frequency \(\omega\).
\( \textbf{Meaning of angular frequency:} \) It is the rate of change of phase with time.
\[
\omega=\frac{\Delta \theta}{\Delta t}
\]
\( \textbf{Substitution:} \)
\[
\omega=\frac{6\pi\,\text{rad}}{0.30\,\text{s}}
\]
\( \textbf{Calculation:} \)
\[
\omega=20\pi\,\text{rad s}^{-1}
\]
\( \textbf{Final answer:} \) The angular frequency is \(20\pi\,\text{rad s}^{-1}\).
This calculation uses phase change, not distance travelled by the wave.
111. A phase changes by \(2\pi\,\text{rad}\) when position changes by one wavelength. This statement leads to
ⓐ. \(\omega=\frac{2\pi}{T}\)
ⓑ. \(k=\frac{2\pi}{\lambda}\)
ⓒ. \(\Delta\phi=\omega\Delta t\)
ⓓ. \(T=\frac{1}{f}\)
Correct Answer: \(k=\frac{2\pi}{\lambda}\)
Explanation: Wave number \(k\) measures phase change per unit distance. A complete spatial cycle of a sinusoidal wave corresponds to a phase change of \(2\pi\,\text{rad}\). The distance for one complete spatial cycle is the wavelength \(\lambda\). Therefore, phase change per unit distance is \(k=\frac{2\pi}{\lambda}\). The relation \(\omega=\frac{2\pi}{T}\) is similar in form, but it concerns phase change with time. The given statement is spatial, so it points to \(k\), not \(\omega\).
112. Study the table and identify the row that describes the phase quantities properly.
| Row | Quantity | Meaning |
|---|
| P | \(\omega\) | Rate of phase change with time |
| Q | \(k\) | Rate of phase change with position |
| R | \(\phi\) | Constant phase shift |
| S | \(\lambda\) | Number of oscillations per second |
The mismatched row is
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: Row S is mismatched because \(\lambda\) is wavelength, the spatial distance between nearest points in the same phase. The number of oscillations per second is frequency \(f\), not wavelength. Row P is suitable because \(\omega\) gives phase change per unit time. Row Q is suitable because \(k\) gives phase change per unit position. Row R is suitable because \(\phi\) sets a constant shift in phase. The mismatch replaces a spatial quantity with a time-rate quantity.
113. A wave is described by a phase \(kx-\omega t+\phi\). At fixed \(x\), increasing \(t\) changes the phase mainly through the term
ⓐ. \(kx\) only
ⓑ. \(-\omega t\)
ⓒ. \(\phi\) only
ⓓ. \(k\omega t\)
Correct Answer: \(-\omega t\)
Explanation: At fixed \(x\), the value of \(kx\) remains constant because the position is not changing. The phase constant \(\phi\) also remains fixed. The time variation of the phase is therefore controlled by the term involving \(t\), namely \(-\omega t\). The angular frequency \(\omega\) tells how quickly the phase changes with time at that fixed position. The product \(k\omega\) is not a separate term in the given phase expression. This is why a \(y-t\) graph at one position is connected with \(\omega\) and \(T\).
114. A wave has \(\omega=40\pi\,\text{rad s}^{-1}\) and \(k=5\pi\,\text{m}^{-1}\). Its wave speed is
ⓐ. \(4\,\text{m s}^{-1}\)
ⓑ. \(20\,\text{m s}^{-1}\)
ⓒ. \(200\,\text{m s}^{-1}\)
ⓓ. \(8\,\text{m s}^{-1}\)
Correct Answer: \(8\,\text{m s}^{-1}\)
Explanation: \( \textbf{Given data:} \) \(\omega=40\pi\,\text{rad s}^{-1}\), \(k=5\pi\,\text{m}^{-1}\).
\( \textbf{Required quantity:} \) Wave speed \(v\).
\( \textbf{Relation between phase constants and speed:} \)
\[
v=\frac{\omega}{k}
\]
This follows because \(\omega\) measures phase change per unit time and \(k\) measures phase change per unit distance.
\( \textbf{Substitution:} \)
\[
v=\frac{40\pi\,\text{s}^{-1}}{5\pi\,\text{m}^{-1}}
\]
\( \textbf{Cancellation:} \)
\[
\frac{40\pi}{5\pi}=8
\]
\( \textbf{Unit check:} \)
\[
\frac{\text{s}^{-1}}{\text{m}^{-1}}=\text{m s}^{-1}
\]
\( \textbf{Final answer:} \) The wave speed is \(8\,\text{m s}^{-1}\).
The factors of \(\pi\) cancel because both \(\omega\) and \(k\) are phase-rate quantities.
115. The relation \(v=\frac{\omega}{k}\) is consistent with \(v=f\lambda\) because
ⓐ. \(\omega=2\pi f,\ k=\frac{2\pi}{\lambda}\)
ⓑ. \(\omega=\frac{f}{2\pi}\) and \(k=2\pi\lambda\)
ⓒ. \(\omega=f\lambda\) and \(k=\frac{2\pi}{T}\)
ⓓ. \(\omega=\lambda f\) and \(k=\frac{1}{T}\)
Correct Answer: \(\omega=2\pi f,\ k=\frac{2\pi}{\lambda}\)
Explanation: The angular frequency is related to ordinary frequency by \(\omega=2\pi f\). The wave number is related to wavelength by \(k=\frac{2\pi}{\lambda}\). Substituting these in \(v=\frac{\omega}{k}\) gives \(v=\frac{2\pi f}{2\pi/\lambda}\). The factor \(2\pi\) cancels, leaving \(v=f\lambda\). This shows that both forms describe the same wave speed. The relation \(v=\frac{\omega}{k}\) uses phase rates, while \(v=f\lambda\) uses cycles and wavelength.
116. A travelling wave enters a second part of a string where the wave speed becomes smaller, while the source frequency remains unchanged. The wavelength in the second part
ⓐ. increases because \(v=f\lambda\)
ⓑ. decreases because \(f\) stays fixed and \(v\) decreases
ⓒ. remains unchanged because wavelength is fixed only by the source
ⓓ. becomes zero because the wave crosses a boundary
Correct Answer: decreases because \(f\) stays fixed and \(v\) decreases
Explanation: When a wave crosses into a new medium, the source frequency remains the same because the source continues to oscillate at the same rate. The wave speed, however, depends on the medium. The relation \(v=f\lambda\) then shows that if \(v\) decreases while \(f\) stays fixed, \(\lambda\) must decrease. The wavelength adjusts to match the speed in the new medium. It is not fixed only by the source. The frequency is the source-controlled quantity, while the speed is medium-controlled for a mechanical wave.
117. A wave source produces waves of frequency \(12\,\text{Hz}\) in a medium where the wave speed is \(36\,\text{m s}^{-1}\). If the same source produces waves in another medium where the speed is \(24\,\text{m s}^{-1}\), the new wavelength is
ⓐ. \(2.0\,\text{m}\)
ⓑ. \(3.0\,\text{m}\)
ⓒ. \(12\,\text{m}\)
ⓓ. \(24\,\text{m}\)
Correct Answer: \(2.0\,\text{m}\)
Explanation: \( \textbf{Given source frequency:} \) \(f=12\,\text{Hz}=12\,\text{s}^{-1}\).
\( \textbf{New medium speed:} \) \(v_2=24\,\text{m s}^{-1}\).
\( \textbf{Required quantity:} \) New wavelength \(\lambda_2\).
\( \textbf{Relation used:} \)
\[
v=f\lambda
\]
The source frequency is unchanged when the same source drives waves in the second medium.
\( \textbf{Rearranging:} \)
\[
\lambda_2=\frac{v_2}{f}
\]
\( \textbf{Substitution:} \)
\[
\lambda_2=\frac{24\,\text{m s}^{-1}}{12\,\text{s}^{-1}}
\]
\( \textbf{Calculation:} \)
\[
\lambda_2=2.0\,\text{m}
\]
\( \textbf{Final answer:} \) The new wavelength is \(2.0\,\text{m}\).
The earlier speed \(36\,\text{m s}^{-1}\) is useful for comparison, but the new wavelength must use the new medium speed.
118. A periodic wave has frequency \(6.0\,\text{Hz}\) and wavelength \(0.75\,\text{m}\). Another wave in the same medium has frequency \(9.0\,\text{Hz}\). If the medium fixes the speed, the wavelength of the second wave is
ⓐ. \(0.75\,\text{m}\)
ⓑ. \(1.125\,\text{m}\)
ⓒ. \(4.5\,\text{m}\)
ⓓ. \(0.50\,\text{m}\)
Correct Answer: \(0.50\,\text{m}\)
Explanation: \( \textbf{First wave data:} \) \(f_1=6.0\,\text{Hz}\), \(\lambda_1=0.75\,\text{m}\).
\( \textbf{Medium condition:} \) Both waves are in the same medium, so the wave speed is the same.
\( \textbf{Speed from first wave:} \)
\[
v=f_1\lambda_1
\]
\( \textbf{Substitution:} \)
\[
v=(6.0\,\text{s}^{-1})(0.75\,\text{m})=4.5\,\text{m s}^{-1}
\]
\( \textbf{Second wave frequency:} \) \(f_2=9.0\,\text{Hz}\).
\( \textbf{Wavelength of second wave:} \)
\[
\lambda_2=\frac{v}{f_2}
\]
\( \textbf{Substitution:} \)
\[
\lambda_2=\frac{4.5\,\text{m s}^{-1}}{9.0\,\text{s}^{-1}}
\]
\( \textbf{Calculation:} \)
\[
\lambda_2=0.50\,\text{m}
\]
\( \textbf{Final answer:} \) The wavelength of the second wave is \(0.50\,\text{m}\).
In the same medium, a higher frequency corresponds to a shorter wavelength.
119. A table compares two cases for waves in the same medium.
| Case | Frequency | Wavelength |
|---|
| P | \(f\) | \(\lambda\) |
| Q | \(2f\) | \(\frac{\lambda}{2}\) |
The table is based on the condition that
ⓐ. wave speed remains the same
ⓑ. wave speed becomes double
ⓒ. time period remains the same
ⓓ. wavelength is independent of frequency in the same medium
Correct Answer: wave speed remains the same
Explanation: For a wave in the same medium, the propagation speed is usually fixed by the medium conditions. The relation \(v=f\lambda\) must then remain constant. If frequency becomes \(2f\), the wavelength must become \(\frac{\lambda}{2}\) so that the product \(f\lambda\) is unchanged. The time period does not remain the same, because doubling frequency halves the time period. The table therefore represents inverse change of \(f\) and \(\lambda\) at constant \(v\). It does not show wavelength being independent of frequency.
120. Assertion: For mechanical waves in a given uniform medium, changing the source frequency does not by itself change the wave speed.
Reason: In a given medium, the wave speed is determined by the medium’s physical properties.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The Assertion is true for the usual school-level treatment of mechanical waves in a fixed uniform medium. Changing the source frequency changes how often wave crests or compressions are produced. The wave speed is controlled by the medium’s properties, such as tension and linear mass density for a string. If the speed remains fixed, the wavelength adjusts according to \(v=f\lambda\). The Reason correctly states why frequency change alone does not decide the speed in a given medium. The source controls \(f\), while the medium controls \(v\).
