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Physics M.C.Q (1 Marks)

Wave Motion and Waves on a String · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 32 questions.

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M.C.Q (1 Marks)

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32 questions · 7 auto-graded MCQ + 25 self-marked written.

Question 11 Mark
The fundamental frequency of a string is proportional to:
  1. Inverse of its length.
  2. The diameter.
  3. The tension.
  4. The density.
Answer
  1. Inverse of its length.
Explanation:
The relation between wave speed and the length of the string is given by
$\text{v}=\frac{1}{2\text{l}}\sqrt{\frac{\text{F}}{\mu}}$
where
l is the length of the string
F is the tension
$\mu$ linear mass density
From the above relation, we can say that the fundamental frequency of a string is proportional to the inverse of the length of the string.
$\text{v}\propto\frac{1}{\text{l}}.$
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Question 21 Mark
Velocity of sound in air is 332m/s. Its velocity in vacuum will be:
  1. > 332m/s
  2. = 332 m/s
  3. < 332m/s
  4. Meaningless.
Answer
  1. Meaningless.
Explanation:
Sound wave is a mechanical wave; this means that it needs a medium to travel. Thus, its velocity in vacuum is meaningless.
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Question 31 Mark
Consider two waves passing through the same string. Principle of superposition for displacement says that the net displacement of a particle on the string is sum of the displacements produced by the two waves individually. Suppose we state similar principles for the net velocity of the particle and the net kinetic energy of the particle. Such a principle will be valid for:
  1. Both the velocity and the kinetic energy.
  2. The velocity but not for the kinetic energy.
  3. The kinetic energy but not for the velocity.
  4. Neither the velocity nor the kinetic energy.
Answer
  1. The velocity but not for the kinetic energy.
Explanation:
The principle of superposition Is valid only for vector quantities. Velocity is a vector quantity, but kinetic energy is a scalar quantity.
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Question 41 Mark
A cork floating in a calm pond executes simple harmonic motion of frequency v when a wave generated by a boat passes by it. The frequency of the wave is:
  1. $\text{v}$
  2. $\frac{\text{v}}{2}$
  3. $2\text{v}$
  4. $\sqrt{2}\text{v}.$
Answer
  1. $\text{v}$
Explanation:
The boat transmits the same wave without any change of frequency to cause the cork to execute SHM with same frequency though amplitude may differ.
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Question 51 Mark
A wave moving in a gas,
  1. Must be longitudinal
  2. May be longitudinal
  3. Must be transverse
  4. May be transverse
Answer
  1. Must be longitudinal.
Explanation:
Because particles in a gas are far apart, only longitudinal wave can travel through it.
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Question 61 Mark
The equation $\text{y}=\text{A}\sin^2(\text{kx}-\omega\text{t})$ represents a wave motion with:
  1. Amplitude A, frequency $\frac{\omega}{2\pi}$
  2. Amplitude $\frac{\text{A}}{2},$ frequency $\frac{\omega}{\pi}$
  3. Amplitude 2A, frequency $\frac{\omega}{4\pi}$
  4. Does not represent a wave motion.
Answer
  1. Amplitude $\frac{\text{A}}{2},$ frequency $\frac{\omega}{\pi}$
Explanation:
$\text{y}=\text{A}\sin^2(\text{kx}-\omega\text{t})$
$\Big[\cos^2\theta=1-2\sin^2\theta,\ \sin^2\theta=\frac{1-\cos^2\theta}{2}\Big]$
$\text{y}=\text{A}\Big[\frac{1-\cos^2(\text{kx}-\omega\text{t})}{2}\Big]$
$\text{y}=\frac{\text{A}}{2}\Big[1-\cos^2(\text{kx}-\omega\text{t})\Big]$
Thus, we have:
Amplitude $=\frac{\text{A}}{2}$
Frequency $=2\Big(\frac{\omega}{2\pi}\Big)$
$=\frac{\omega}{\pi}$
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Question 71 Mark
Which of the following is a mechanical wave?
  1. Radio waves.
  2. X-rays.
  3. Light waves.
  4. Sound waves.
Answer
  1. Sound waves.
Explanation:
There are mainly two types of waves: first is electromagnetic wave, which does not require any medium to travel, and the second is the mechanical wave, which requires a medium to travel. sound requires medium to travel. hence it is a mechanical wave.
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Question 81 Mark
A mechanical wave propagates in a medium along the X-axis. The particles of the medium.
  1. Must move on the X-axis
  2. Must move on the Y-axis
  3. May move on the X-axis
  4. May move on the Y-axis.
Answer
  1. May move on the X-axis
  2. May move on the Y-axis.
Explanation:
A mechanical wave is of two types: longitudinal and transverse. So, a particle of a mechanical wave may move perpendicular or along the direction of motion of the wave.
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Question 91 Mark
A wave going in a solid.
  1. Must be longitudinal
  2. May be longitudinal
  3. Must be transverse
  4. May be transverse.
Answer
  1. May be longitudinal
  1. May be transverse.
Explanation:
Particles in a solid are very close to each other; thus. both longitudinal and transverse waves can travel through it.
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Question 101 Mark
A transverse wave travels along the Z-axis. The particles of the medium must move:
  1. Along the Z-axis
  2. Along the X-axis
  3. Along the Y-axis
  4. In the X-Y plane.
Answer
  1. In the X-Y plane.
Explanation:
In a transverse wave. particles move perpendicular to the direction of motion of the wave. In other words, if a wave moves along the Z-axis, the particles will move in the X-Y plane.
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Question 111 Mark
Mark out the correct options.
  1. The energy of any small part of a string remains constant in a travelling wave.
  2. The energy of any small part of a string remains constant in a standing wave. 
  3. The energies of all the small parts of equal length are equal in a travelling wave.
  4. The energies of all the small parts of equal length are equal in a standing wave.
Answer
  1. The energy of any small part of a string remains constant in a standing wave. 
Explanation:
A standing wave is formed when the energy of any small part of a string remains constant. If it does not, then there is transfer of energy. In that case. the wave is not stationary.
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Question 121 Mark
Longitudinal waves cannot:
  1. Have a unique wavelength
  2. Transmit energy
  3. Have a unique wave velocity
  4. Be polarized.
Answer
  1. Be polarized.
Explanation:
A longitudinal wave has particle displacement along its direction of motion; thus, it cannot be polarised.
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Question 131 Mark
Two wave pulses travel in opposite directions on a string and approach each other. The shape of one pulse is inverted with respect to the other.
  1. The pulses will collide with each other and vanish after collision.
  2. The pulses will reflect from each other i.e., the pulse going towards right will finally move towards left and vice versa.
  3. The pulses will pass through each other but their shapes will be modified.
  4. The pulses will pass through each other without any change in their shapes.
Answer
  1. The pulses will pass through each other without any change in their shapes.
Explanation:
The pulses continue to retain their Identity after they meet, but the moment they meet their wave profile differs from the individual pulse.
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Question 141 Mark
In a stationary wave,
  1. All the particles of the medium vibrate in phase.
  2. All the antinodes vibrate in phase.
  3. The alternate antinodes vibrate in phase.
  4. All the particles between consecutive nodes vibrate in phase.
Answer
  1. The alternate antinodes vibrate in phase.
  1. All the particles between consecutive nodes vibrate in phase.
Explanation:
All particles in a particular segment between two nodes vibrate in the same phase, but the particles in the neighbouring segments vibrate in opposite phases, as shown below.

Thus, particles in alternate antinodes vibrate in the same phase.
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Question 151 Mark
A tuning fork of frequency 480Hz is used to vibrate a sonometer wire having natural frequency 410Hz. The wire will vibrate with a frequency.
  1. 410Hz
  2. 480Hz
  3. 820Hz
  4. 960Hz.
Answer
  1. 480Hz
Explanation:
The frequency of vibration of a sonometer wire Is the same as that of a fork. If this happens to be the natural frequency of the wire, standing waves with large amplitude are set In It.
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Question 161 Mark
A tuning fork of frequency 480Hz is used to vibrate a sonometer wire having natural frequency 240Hz. The wire will vibrate with a frequency of:
  1. 240Hz
  2. 480Hz
  3. 720Hz
  4. Will not vibrate.
Answer
  1. 480Hz
Explanation:
The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be natural frequency of the wire, then standing waves with large amplitude are set up in it.
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MCQ 171 Mark
Two periodic waves of amplitudes $A_1$ and $A_2$ pass through a region. If $A_1 > A_2$, the difference in the maximum and minimum resultant amplitude possible is:
  • A
    $2A_1$
  • ✓
    $2A_2$
  • C
    $A_1 + A_2$
  • D
    $A_1 - A_2$
Answer
Correct option: B.
$2A_2$
we know resultant amplitude is given by
$\text{A}_\text{net}=\sqrt{\text{A}_1^2+\text{A}_2^2+2\text{A}_1\text{A}_2\cos\phi}$
For maximum resultant amplitude
$\text{A}_\text{max}=\text{A}_1+\text{A}_2$
For mnimum resultant amplitude
$\text{A}_\text{min}=\text{A}_1-\text{A}_2$
So, the difference between $A_\text{max}$ and $A_{\min}$ is
$\text{A}_\text{max}-\text{A}_\text{min}=\text{A}_1+\text{A}_2-\text{A}_1+\text{A}_2=2\text{A}_2$
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MCQ 181 Mark
A sonometer wire of length l vibrates in fundamental mode when excited by a tuning fork of frequency $416\ Hz$. If the length is doubled keeping other things same, the string will:
  • A
    Vibrate with a frequency of $416\ Hz$
  • ✓
    Vibrate with a frequency of $208\ Hz$
  • C
    Vibrate with a frequency of $832\ Hz$
  • D
    Stop vibrating.
Answer
Correct option: B.
Vibrate with a frequency of $208\ Hz$
According to the relation of the fundamental frequency of a string
$\nu=\frac{1}{2\text{l}}\sqrt{\frac{\text{F}}{\mu}}$
Where
$l$ Is the length of the string
$F$ Is the tension
$\mu$ Is the linear mass density
We know that $v_1 = 416Hz, l_1 = l$ and $l_2 = 2l.$
$\text{v}_1\propto\frac{1}{\text{l}_1}$
$\text{v}_1\text{l}_1=\text{v}_2\text{l}_2$
$(416)\text{l}=\text{v}_2(2\text{l})$
$\text{v}_2=208\text{Hz}$
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MCQ 191 Mark
Two waves of equal amplitude A, and equal frequency travel in the same direction in a medium. The amplitude of the resultant wave is:
  • ✓
    $0$
  • B
    $A$
  • C
    $2A$
  • D
    Between $0$ and $2A$
Answer
Correct option: A.
$0$
The amplitude of the resultant wave depends on the way two waves superimpose,
i.e., the phase angle $\phi.$
So, the resultant amplitude lies between the maximum resultant amplitude $(A_{\max})$ and the minimum resultant amplitude
​​​​​​​$(A_{\min}).A_{\max} = A + A = 2A$
$A_{\min} = A - A = 0$
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MCQ 201 Mark
A sonometer wire supports a $4\ kg$ load and vibrates in fundamental mode with a tuning fork of frequency $416\ Hz.$ The length of the wire between the bridges is now doubled. In order to maintain fundamental mode, the load should be changed to:
  • A
    $1\ kg$
  • B
    $2\ kg$
  • C
    $8\ kg$
  • ✓
    $16\ kg.$
Answer
Correct option: D.
$16\ kg.$
According to the relation of the fundamental frequency of a string $\nu=\frac{1}{2\text{l}}\sqrt{\frac{\text{F}}{\mu}}$
Where $l$ is the length of the string
$F$ is the tension
$\mu$ is the linear mass density of the string
We know that $v_1 = 416\ Hz, l_1 = l$ and $l_2 = 2l$
Also, $m_1 = 4\ kg$ and $m_2 =$ ?
$\nu_1=\frac{1}{2\text{l}_1}\sqrt{\frac{\text{m}_1\text{g}}{\mu}}\ \dots(1)$
$\nu_2=\frac{1}{2\text{l}_2}\sqrt{\frac{\text{m}_2\text{g}}{\mu}}\ \dots(2)$
So, in order to maintain the same fundamental mode
$\nu_1=\nu_2$
squaring both sides of equations $(1)$ and $(2)$ and then equating
$\frac{1}{4\text{l}^2}\frac{4\text{g}}{\mu}=\frac{1}{16\text{l}^2}\frac{\text{m}_2\text{g}}{\mu}$
$\Rightarrow\text{m}_2=16\text{  kg}$
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Question 211 Mark
A wave pulse, travelling on a two -piece string, gets partially reflected and partially transmitted at the junction. The reflected wave is inverted in shape as compared to the incident one. If the incident wave has wavelengt $\lambda$ and the transmitted wave $\lambda'.$
  1. $\lambda'>\lambda$
  2. $\lambda'=\lambda$
  3. $\lambda'<\lambda$
  4. Nothing can be said about the relation of $\lambda$ and $\lambda'.$
Answer
  1. $\lambda'<\lambda$
Explanation:

As $\text{v}=\sqrt{\frac{\text{f}}{\mu}}$
A wave pulse travels faster in a thinner string. The wavelength of the transmitted wave is equal to the wavelength of the Incident wave because the frequency remains constant.
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Question 221 Mark
A wave is represented by the equation-$\text{y}=(0.001\text{mm})\sin\Big[(50\text{s}^{-1})\text{t}+(2.0\text{m}^{-1})\text{x}\Big]$
  1. The wave velocity = 100m/s.
  2. The wavelength = 2.0m.
  3. The frequency $=\frac{25}{\pi}\text{Hz}$
  4. The amplitude = 0.001mm.
Answer
  1. The frequency $=\frac{25}{\pi}\text{Hz}$
  1. The amplitude = 0.001mm.
Explanation:
$\text{y}=(0.001\text{mm})\sin\Big[(50\text{s}^{-1})\text{t}+(2.0\text{m}^{-1})\text{x}\Big]$
Equating the above equation with the general equation, we get:
$\text{y}=\text{A}\sin(\omega\text{t}-\text{kx})$
$\omega=\frac{2\pi}{\text{T}}=2\pi\nu$
$\text{k}=\frac{2\pi}{\lambda}$
Here. A is the amplitude, $\omega$ is the angular frequency, k is the wave number and $\lambda$ is the wavelength.
$\text{A}=0.001\text{mm}$
Now,
$50=2\pi\nu$
$\Rightarrow\nu=\frac{25}{\pi}\text{Hz}$
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Question 231 Mark
Two particles A and B have a phase difference of $\pi$ when a sine wave passes through the region.
  1. A oscillates at half the frequency of B.
  2. A and B move in opposite directions.
  3. A and B must be separated by half of the wavelength.
  4. The displacements at A and B have equal magnitudes.
Answer
  1. A and B move in opposite directions.
  1. The displacements at A and B have equal magnitudes.
Explanation:

A and Bhave a phase difference of fl. So, when a sine wave passes through the region, they move in opposite directions and have equal displacement. They may be separated by any odd multiple of their wavelength.
$\overrightarrow{\text{yA}}=\text{A}\sin(\omega\text{t})$
$\overrightarrow{\text{yB}}=\text{B}\sin(\omega\text{t}+\pi)$
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Question 241 Mark
A sine wave is travelling in a medium. A particular particle has zero displacement at a certain instant. The particle closest to it having zero displacement is at a distance:
  1. $\frac{\lambda}{4}$
  2. $\frac{\lambda}{3}$
  3. $\frac{\lambda}{2}$
  4. $\lambda.$
Answer
  1. $\frac{\lambda}{2}$
Explanation:
A sine wave has a maxima and a minima and the particle displacement has phase difference of $\pi$ radians. Therefore, applying similar argument we can say that if a particular particle has zero displacement at a certain instant. then the particle closest to it having zero displacement is at a distance is equal to $\frac{\lambda}{2}.$
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Question 251 Mark
A sine wave is travelling in a medium. The minimum distance between the two particles, always having same speed, is:
  1. $\frac{\lambda}{4}$
  2. $\frac{\lambda}{3}$
  3. $\frac{\lambda}{2}$
  4. $\lambda.$
Answer
  1. $\frac{\lambda}{2}$
 Explanation:
A sme wave has a maxima and a minimum and the particle displacement has phase difference of n radians. The speeds at the maximum point and at the minimum point are same although the direction of motion are different. The difference between the posiuons of maxima and minima Is equal to $\frac{\lambda}{2}.$
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Question 261 Mark
Two strings A and B, made of same material, are stretched by same tension. The radius of string A is double of the radius of B. A transverse wave travels on A with speed $\nu_\text{A}$ and on B with speed $\nu_\text{B}.$ The ratio $\frac{\nu_\text{A}}{\nu_\text{B}}$ is:
  1. $\frac{1}{2}$
  2. $2$
  3. $\frac{1}{4}$
  4. $4.$
Answer
  1. $\frac{1}{2}$
Explanation:
wave speed is given by $\nu=\sqrt{\frac{\text{T}}{\mu}}$
Where
T is the tension in the string
v is the speed of the wave
$\mu$ Is the mass per unit length of the stnng
$\mu=\frac{\text{M}}{\text{L}}=\rho\frac{\text{V}}{\text{L}}=\rho\frac{(\text{AL})}{\text{L}}$
Where
Mis the mass of the stnng. which can be written as pV
L Is the length of the string
$=\rho(\pi\text{r}^2)=\rho\Big(\pi\frac{\text{D}^2}{4}\Big)$
$\therefore\nu=\sqrt{\frac{\text{T}}{\rho\pi\frac{\text{D}^2}{4}}}=\frac{2}{\text{D}}\sqrt{\frac{\text{T}}{\rho\pi}}$
Where Dis the diameter of the string.
Thus, $\text{V}\propto\frac{1}{\text{D}}$
Since, $\text{r}_\text{A}=2\text{r}_\text{B}$
$\text{v}_\text{A}\propto\frac{1}{2\text{r}_\text{A}}\propto\frac{1}{2\times2\text{r}_\text{B}}\ \dots(1)$
$\text{v}_\text{B}\propto\frac{1}{2\text{r}_\text{B}}\ \dots(2)$
From Equations (1) and (2) we get
$\frac{\text{v}_\text{A}}{\text{v}_\text{B}}=\frac{1}{2}$
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Question 271 Mark
A standing wave is produced on a string clamped at one end and free at the other. The length of the string.
  1. Must be an integral multiple of $\frac{\lambda}{4}$
  2. Must be an integral multiple of $\frac{\lambda}{2}$
  3. Must be an integral multiple of $\lambda$
  4. May be an integral multiple of $\frac{\lambda}{2}.$
Answer
  1. Must be an integral multiple of $\frac{\lambda}{4}$
Explanation:
A standing wave is produced on a string clamped at one end and free at the other. Its fundamental frequency is given by
$\nu=\Big(\text{n}+\frac{1}{2}\Big)\frac{\nu}{2\text{L}}$
$\Rightarrow\text{v}=\nu\lambda$
$\Rightarrow\nu=\Big(\text{n}+\frac{1}{2}\Big)\frac{\nu\lambda}{2\text{L}}$
$\Rightarrow\text{L}=\Big(\frac{2\text{n}+1}{4}\Big)\lambda$
$\Rightarrow\text{L}=\frac{\lambda}{4},\frac{3\lambda}{4},\ \dots$
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Question 281 Mark
Which of the following equations represents a wave travelling along Y-axis?
  1. $\text{x}=\text{A}\sin(\text{ky}-\omega\text{t})$
  2. $\text{y}=\text{A}\sin(\text{kx}-\omega\text{t})$
  3. $\text{y}=\text{A}\sin\text{ky}\cos\ \omega\text{t}$
  4. $\text{y}=\text{A}\cos\text{ky}\sin\ \omega\text{t}.$
Answer
  1. $\text{x}=\text{A}\sin(\text{ky}-\omega\text{t})$
Explanation:
Here x is the particle displacement of the wave and the wave is travelling along the Y-axis because the particle displacement is perpendicular to the direction of wave motion.
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Question 291 Mark
Two waves represented by $\text{y}=\text{a}\sin(\omega\text{t}-\text{kx})$ and $\text{y}=\text{a}\cos(\omega\text{t}-\text{kx})$ are superposed. The resultant wave will have an amplitude:
  1. a
  2. $\sqrt{2}\text{a}$
  3. 2a
  4. 0.
Answer
  1. $\sqrt{2}\text{a}$
Explanation:
We know that the resultant of the amplitude is given by
$\text{R}_\text{net}=\sqrt{\text{A}_1^2+\text{A}_2^2+2\text{A}_1\text{A}_2\cos\phi}$
For the particular case, we can write
$=\sqrt{\text{a}^2+\text{a}^2+2\text{a}^2\cos\frac{\pi}{2}}$
$=\sqrt{2}\text{a}$
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MCQ 301 Mark
Two wires $A$ and $B,$ having identical geometrical construction, are stretched from their natural length by small but equal amount. The Young's modulus of the wires are $Y_A$ and $Y_B$ whereas the densities are $\rho_\text{A}>\rho_{\text{B}^+}$ It is given that $Y_A > Y_B$ and $\rho_\text{A}>\rho_{\text{B}^+}.$ A transverse signal started at one end takes a time $t_1$ to reach the other end for $A$ and $t_2$ for $B.$
  • A
    $t_1 < t_2$
  • B
    $t_1 = t_2$
  • C
    $t_1 > t_2$
  • ✓
    The information is insufficient to find the relation betwee $t_1$ and $t_2.$
Answer
Correct option: D.
The information is insufficient to find the relation betwee $t_1$ and $t_2.$
$\text{v}=\sqrt{\frac{\eta}{\rho}}$
But because the length of wires $A$ and $B$ is not known, the relation between $A$ and $B$ cannot be determined.
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MCQ 311 Mark
Both the strings, shown in figure are made of same material and have same cross$-$section. The pulleys are light. The wave speed of a transverse wave in the string $AB$ is $\nu_1$ and in $CD$ it is $\nu_2.$ Then $\frac{\nu_1}{\nu_2}$ is:
  • A
    $1$
  • B
    $2$
  • C
    $\sqrt{2}$
  • ✓
    $\frac{1}{\sqrt{2}}.$
Answer
Correct option: D.
$\frac{1}{\sqrt{2}}.$

$T_{AB} = T$
$T_{CD} = 2T$
where
$T_{AB}$ is the tension in the string $AB$
$T_{CD}$ is the tension in the stnng $CD$
The eelatlon between tension and the wave speed is given by
$\text{v}=\sqrt{\frac{\text{T}}{\mu}}$
$\text{v}\propto\sqrt{\text{T}}$
where
$v$ is the wave speed of the transverse wave
$\mu$ is the mass per unit length of the string
$\frac{\text{v}_1}{\text{v}_2}=\sqrt{\frac{\text{T}}{2\text{T}}}$
$=\frac{1}{\sqrt{2}}$
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MCQ 321 Mark
Two sine waves travel in the same direction in a medium. The amplitude of each wave is $A$ and the phase difference between the two waves is $120^\circ$ . The resultant amplitude will be:
  • ✓
    $A$
  • B
    $2A$
  • C
    $4A$
  • D
    $\sqrt{2}\text{A}.$
Answer
Correct option: A.
$A$
We know the resultant amplitude Is given by
$\text{R}_\text{net}=\sqrt{\text{A}^2+\text{A}^2+2\text{A}^2\cos120^\circ}$
$(\phi=120^\circ)$
$=\sqrt{2\text{A}^2-\text{A}^2}$
$\Big[\because\cos120^\circ=\frac{-1}{2}\Big]$
$=\text{A}$
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