Q.A sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If the tension in the wire remains the same, by what factor will the frequency change?

Showing the 12 most recent of 23 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.A wire of length 1.2 m is subjected to a tension of 240 N. If the frequencies of two successive modes of vibration of the wire are 200 Hz and 250 Hz, then the mass of the wire is (A) 10 g (B) 20 g (C) 30 g (D) 40 g

   ›Reveal solutionSolution

   The difference between two successive mode frequencies gives the fundamental frequency, which together with the tension and length yields the linear mass density and hence the total mass of the wire. The mass is 20 g.

   The key idea is that for a stretched string fixed at both ends, the allowed vibration frequencies are integer multiples of the fundamental frequency $f_1$. So if you know two successive mode frequencies, their difference is exactly $f_1$. Once you have $f_1$, the wave speed $v$ follows from $v = f_1 \cdot 2L$ (since the fundamental wavelength is $2L$). And the wave speed on a string also equals $\sqrt{T/\mu}$, where $T$ is tension and $\mu$ is mass per unit length. Equating these gives $\mu$, and multiplying by length $L$ gives the total mass.

   Let’s work it through.

   1. Find the fundamental frequency. For a string fixed at both ends, the $n$th harmonic frequency is $f_n = n f_1$. Two successive modes are $f_n = n f_1$ and $f_{n+1} = (n+1)f_1$. Their difference is

   $f_{n+1} - f_n = f_1.$

   Here the given successive frequencies are 200 Hz and 250 Hz, so

   $f_1 = 250 - 200 = 50 \text{ Hz}.$

   2. Relate fundamental frequency to wave speed. For the fundamental mode, the string vibrates with a single loop: wavelength $\lambda_1 = 2L$, where $L = 1.2$ m. The wave speed is

   $v = f_1 \lambda_1 = 50 \times (2 \times 1.2) = 50 \times 2.4 = 120 \text{ m/s}.$

   3. Relate wave speed to tension and linear density. The speed of a transverse wave on a string under tension $T$ is

   $v = \sqrt{\frac{T}{\mu}},$

   where $\mu$ is the mass per unit length. Rearranging,

   $\mu = \frac{T}{v^2}.$

   Substitute $T = 240$ N and $v = 120$ m/s: …

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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.In Young's double slit experiment, if the distance between the first dark fringe and the second bright fringe on the same side of the central maximum is $1.5\ \text{mm}$, then the distance between first dark fringe on one side and the second bright fringe on the other side of the central maximum is (A) $2.5\ \text{mm}$ (B) $1.5\ \text{mm}$ (C) $3.0\ \text{mm}$ (D) $2.0\ \text{mm}$

   ›Reveal solutionSolution

   The key idea is to express the positions of dark and bright fringes in terms of the fringe width $\beta$, then use the given distance to find $\beta$, and finally compute the required distance. The answer is $3.0\ \text{mm}$.

   In Young’s double slit experiment, the pattern on the screen is a series of alternating bright and dark fringes. The central maximum is bright, and on either side, the fringes are equally spaced. The distance between two consecutive bright fringes (or two consecutive dark fringes) is called the fringe width, denoted by $\beta$.

   The position of the $n$th bright fringe (from the central maximum) is given by $y_n = n\beta$, where $n = 0, 1, 2, \dots$ (with $n=0$ being the central maximum). The position of the $n$th dark fringe is given by $y_n' = \left(n + \frac12\right)\beta$, where $n = 0, 1, 2, \dots$ (the first dark fringe corresponds to $n=0$).

   Now, let’s work through the problem step by step.

   1. Identify the given distance.

      The distance between the first dark fringe and the second bright fringe on the same side of the central maximum is $1.5\ \text{mm}$.

      • First dark fringe on one side: $y_0' = \left(0 + \frac12\right)\beta = \frac{\beta}{2}$.
      • Second bright fringe on the same side: $y_2 = 2\beta$ (since $n=2$ for the second bright fringe; the first bright fringe is $n=1$). The distance between them is $2\beta - \frac{\beta}{2} = \frac{3\beta}{2}$. So, $\frac{3\beta}{2} = 1.5\ \text{mm}$, which gives $\beta = 1.0\ \text{mm}$.

   2. Now find the required distance.

      We need the distance between the first dark fringe on one side and the second bright fringe on the other side of the central maximum.

      • First dark fringe on one side: $+\frac{\beta}{2}$ (say to the right). …

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3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.A particle executes simple harmonic motion with an amplitude of 40 cm. If the time period of the particle is 6 s, then the minimum time taken by the particle to move from mean position to a point at a distance of 20 cm from the extreme position is (A) $0.5\ \mathrm{s}$ (B) $1.5\ \mathrm{s}$ (C) $3\ \mathrm{s}$ (D) $2.5\ \mathrm{s}$

   ›Reveal solutionSolution

   A point $20\,\mathrm{cm}$ from the extreme is $20\,\mathrm{cm}$ from the mean, i.e. $x=A/2$; from $x=A\sin\omega t$ this first occurs at $t=0.5\,\mathrm{s}$ — option (A).

   Locate the point. Amplitude $A=40\,\mathrm{cm}$, so the extreme is $40\,\mathrm{cm}$ from the mean. A point $20\,\mathrm{cm}$ from the extreme is

   $40-20 = 20\,\mathrm{cm}\ \text{from the mean}\ \Rightarrow\ x = \tfrac{A}{2}.$

   Angular frequency.

   $\omega = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3}\ \mathrm{rad/s}.$

   Time from the mean. Starting from the mean, $x=A\sin\omega t$: …

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4. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.A wire of length 100 cm is clamped between two rigid supports and is made to vibrate in its fundamental mode. If the amplitude at the midpoint of the wire is $A$, then the distance between two points having amplitude of $\dfrac{A}{\sqrt{2}}$ is (A) 50 cm (B) 60 cm (C) 40 cm (D) 25 cm

   ›Reveal solutionSolution

   For a wire vibrating in its fundamental mode, the displacement profile is a sine wave with a node at each end and an antinode at the centre. The amplitude at a distance $x$ from one end is $A \sin(\pi x / L)$. Setting this equal to $A/\sqrt{2}$ gives $x = L/4$ and $3L/4$, so the distance between these two points is $L/2 = 50$ cm.

   The key idea is that the standing wave on a wire fixed at both ends has a well-defined shape. In the fundamental mode, the wire vibrates as a single loop — the midpoint is the antinode (maximum amplitude $A$), and the ends are nodes (zero amplitude). The amplitude at any other point is simply $A$ times the sine of the distance from a node, scaled to the half-wavelength.

   Let’s work through it.

   1. Set up the standing wave equation. For a wire of length $L = 100$ cm fixed at both ends, the fundamental mode has wavelength $\lambda = 2L = 200$ cm. The displacement at a point a distance $x$ from one end is

   $y(x,t) = A \sin\left(\frac{\pi x}{L}\right) \cos(\omega t)$

   The amplitude of oscillation at that point is the coefficient of the time-dependent part:

   $\text{Amplitude}(x) = A \sin\left(\frac{\pi x}{L}\right)$

   2. Find where the amplitude equals $A/\sqrt{2}$. Set

   $A \sin\left(\frac{\pi x}{L}\right) = \frac{A}{\sqrt{2}}$

   Cancel $A$ (non-zero):

   $\sin\left(\frac{\pi x}{L}\right) = \frac{1}{\sqrt{2}}$

   The principal solutions in $[0, L]$ are

   $\frac{\pi x}{L} = \frac{\pi}{4} \quad \text{and} \quad \frac{\pi x}{L} = \frac{3\pi}{4}$

   because $\sin(\pi/4) = \sin(3\pi/4) = 1/\sqrt{2}$.

   3. Solve for $x$. …

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5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The speed of a progressive transverse wave on a string is $18\,\mathrm{ms}^{-1}$. If the phase difference between two points on the string separated by a distance of $3\,\mathrm{cm}$ is $90^\circ$, then the frequency of the transverse waves on the string is (A) 300 Hz (B) 225 Hz (C) 75 Hz (D) 150 Hz

   ›Reveal solutionSolution

   The key idea is that phase difference relates to path difference via the wavelength, and wave speed equals frequency times wavelength. Using the given phase difference of $90^\circ$ for a $3\,\text{cm}$ separation, we find the wavelength and then the frequency: $f = 150\,\text{Hz}$, so option (D) is correct.

   ---

   The problem gives us the wave speed and the phase difference between two points a known distance apart. The natural link is that phase difference corresponds to a fraction of a full cycle, which in turn corresponds to a fraction of the wavelength. Once we have the wavelength, the frequency follows directly from the wave equation.

   Why this works:

   A wave's phase changes by $360^\circ$ (or $2\pi$ radians) over one full wavelength. So if two points are separated by a distance $\Delta x$, the phase difference $\Delta \phi$ is simply the fraction of a wavelength they are apart, multiplied by $360^\circ$. This gives us a direct proportion:

   $\frac{\Delta \phi}{360^\circ} = \frac{\Delta x}{\lambda}.$

   Then we use $v = f\lambda$ to find $f$.

   ---

   Step-by-step solution:

   1. Convert the phase difference to a fraction of a cycle. A phase difference of $90^\circ$ is one-quarter of a full cycle ($360^\circ$). So

   $\frac{\Delta \phi}{360^\circ} = \frac{90^\circ}{360^\circ} = \frac{1}{4}.$

   2. Relate this fraction to the distance between the points. The distance $\Delta x = 3\,\text{cm} = 0.03\,\text{m}$ corresponds to that same fraction of the wavelength $\lambda$:

   $\frac{\Delta x}{\lambda} = \frac{1}{4}.$

   Hence,

   $\lambda = 4 \times \Delta x = 4 \times 0.03\,\text{m} = 0.12\,\text{m}.$ …

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6. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Two wires A and B made of same material have equal lengths. If the volumes of the wires A and B are in the ratio 1:8 and the tensions applied to the wires A and B are in the ratio 1:2, then the ratio of the speeds of the transverse waves in the wires A and B is (A) $1 : 1$ (B) $2 : 1$ (C) $4 : 1$ (D) $8 : 1$

   ›Reveal solutionSolution

   The speed of a transverse wave in a wire depends on tension and linear mass density. Using volume and length to relate cross‑sectional area and mass, the ratio of speeds comes out to $1:1$.

   The key idea is that the wave speed in a stretched string is $v = \sqrt{T/\mu}$, where $T$ is the tension and $\mu$ is the mass per unit length. Since both wires are made of the same material and have equal lengths, their volumes tell us about their cross‑sectional areas, and hence about their masses. The ratio of tensions is given directly. So we just need to combine these pieces carefully.

   1. Relate volume to cross‑sectional area. For a wire of length $L$ and cross‑sectional area $A$, volume $V = A L$. Since $L$ is the same for both wires, the ratio of areas is the same as the ratio of volumes:

   $\frac{A_A}{A_B} = \frac{V_A}{V_B} = \frac{1}{8}.$

   2. Find the ratio of masses. Both wires are made of the same material, so density $\rho$ is identical. Mass $m = \rho V$, hence

   $\frac{m_A}{m_B} = \frac{V_A}{V_B} = \frac{1}{8}.$

   3. Find the ratio of linear mass densities. Linear mass density $\mu = m/L$. Since $L$ is equal,

   $\frac{\mu_A}{\mu_B} = \frac{m_A}{m_B} = \frac{1}{8}.$

   4. Apply the wave speed formula. Speed $v = \sqrt{T/\mu}$. So

   $\frac{v_A}{v_B} = \sqrt{ \frac{T_A}{T_B} \cdot \frac{\mu_B}{\mu_A} }.$

   Given $T_A : T_B = 1 : 2$, we have $T_A/T_B = 1/2$. And $\mu_B/\mu_A = 8$ (from step 3). Therefore

   $$$$

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7. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.The equation of transverse displacement of a wire of mass 20 g and length 100 cm clamped at its ends is $y(x, t) = 0.05 \sin \left( \frac{2\pi}{5} x \right) \cos (80\pi t)$, where ‘x’ is in metre and ‘t’ is in second. The tension in the wire is (A) 200 N (B) 600 N (C) 400 N (D) 800 N

   ›Reveal solutionSolution

   We determine the wave speed from the given standing wave equation and then use the formula relating wave speed, tension, and linear mass density to find the tension in the wire. The tension in the wire is $\boxed{800 \text{ N}}$.

   When a wire clamped at its ends vibrates, it forms a standing wave. The equation describing this transverse displacement contains information about the wave's characteristics, such as its wave number and angular frequency. These characteristics, in turn, allow us to calculate the speed at which transverse waves propagate along the wire. The speed of a transverse wave on a string is fundamentally determined by the tension in the string and its linear mass density. By extracting the wave speed from the given equation and calculating the linear mass density from the provided mass and length, we can then solve for the tension.

   Here's how we can find the tension:

   1. Identify the standard standing wave equation and extract wave parameters:

      The general equation for a standing wave on a string, formed by two waves travelling in opposite directions, can be written as $y(x, t) = A \sin(kx) \cos(\omega t)$.

      Comparing this with the given equation for the transverse displacement:

      $y(x, t) = 0.05 \sin \left( \frac{2\pi}{5} x \right) \cos (80\pi t)$

      We can directly identify the wave number ($k$) and the angular frequency ($\omega$):

      $k = \frac{2\pi}{5} \text{ rad/m}$

      $\omega = 80\pi \text{ rad/s}$

   2. Calculate the wave speed ($v$):

      The speed of a wave ($v$) is related to its angular frequency ($\omega$) and wave number ($k$) by the formula:

      $v = \frac{\omega}{k}$

      Substituting the values we found:

      $v = \frac{80\pi \text{ rad/s}}{\frac{2\pi}{5} \text{ rad/m}} = \frac{80\pi \times 5}{2\pi} \text{ m/s} = 40 \times 5 \text{ m/s} = 200 \text{ m/s}$

      So, the speed of the transverse wave on the wire is $200 \text{ m/s}$.

   3. Calculate the linear mass density ($\mu$) of the wire: …

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8. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The fundamental frequency of transverse wave of a stretched string subjected to a tension $T_1$ is $300$ Hz. If the length of the string is doubled and subjected to a tension of $T_2$, the fundamental frequency of the transverse wave in the string becomes $100$ Hz, then $T_2 : T_1 =$ (Linear density of the string is constant) (A) $1 : 2$ (B) $3 : 4$ (C) $2 : 3$ (D) $4 : 9$

   ›Reveal solutionSolution

   The fundamental frequency of a stretched string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$. Using the given data for two different tensions and lengths, we set up a ratio and solve for $T_2 : T_1$, obtaining $4 : 9$. The correct option is (D).

   The key concept here is the standing wave on a stretched string. For a string fixed at both ends, the fundamental (lowest) frequency is determined by the length $L$, the tension $T$, and the linear mass density $\mu$ (mass per unit length). The formula is:

   $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$

   This comes from the wave speed $v = \sqrt{T/\mu}$ and the fact that the fundamental wavelength is $2L$. So frequency is inversely proportional to length and proportional to the square root of tension. Since the linear density is constant, we can compare two situations directly.

   1. Write the frequency for the first case. Given: $f_1 = 300$ Hz, tension $T_1$, length $L_1 = L$ (say).

   $300 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}}$

   2. Write the frequency for the second case. Given: $f_2 = 100$ Hz, tension $T_2$, length $L_2 = 2L$ (doubled).

   $100 = \frac{1}{2(2L)} \sqrt{\frac{T_2}{\mu}} = \frac{1}{4L} \sqrt{\frac{T_2}{\mu}}$

   3. Set up the ratio of the two equations. Divide the first equation by the second:

   $\frac{300}{100} = \frac{ \frac{1}{2L} \sqrt{T_1/\mu} }{ \frac{1}{4L} \sqrt{T_2/\mu} }$

   Simplify the left: $300/100 = 3$.

   On the right, the $1/L$ and $\mu$ cancel, and we have:

   $3 = \frac{ \frac{1}{2} \sqrt{T_1} }{ \frac{1}{4} \sqrt{T_2} } = \frac{1/2}{1/4} \cdot \frac{\sqrt{T_1}}{\sqrt{T_2}} = 2 \cdot \frac{\sqrt{T_1}}{\sqrt{T_2}}$ …

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9. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The equation of a transverse wave propagating on a stretched string is given by $y = 3 \sin(4x + 200t)$, where $x$ and $y$ are in metre and the time ‘$t$’ is in second. If the tension applied to the string is $500~\mathrm{N}$, the linear density of the string is (A) $0.25~\mathrm{kg\,m^{-1}}$ (B) $0.4~\mathrm{kg\,m^{-1}}$ (C) $0.2~\mathrm{kg\,m^{-1}}$ (D) $0.1~\mathrm{kg\,m^{-1}}$

   ›Reveal solutionSolution

   The wave speed is found from the wave equation’s angular frequency and wave number, then related to tension and linear density via $v = \sqrt{T/\mu}$. The linear density comes out to $0.2~\mathrm{kg\,m^{-1}}$, so option (C) is correct.

   The key idea is that for a transverse wave on a stretched string, the wave speed is determined by the tension and the linear density: $v = \sqrt{T/\mu}$. But we can also read the wave speed directly from the wave equation $y = A \sin(kx \pm \omega t)$: the speed is $v = \omega/k$. Equating these two expressions lets us solve for the unknown linear density $\mu$.

   ---

   1. Identify the wave parameters from the given equation

      The wave is $y = 3 \sin(4x + 200t)$. Compare with the standard form $y = A \sin(kx + \omega t)$ (the plus sign means the wave travels in the negative $x$-direction, but speed magnitude is what matters).

      • Wave number $k = 4~\mathrm{rad\,m^{-1}}$
      • Angular frequency $\omega = 200~\mathrm{rad\,s^{-1}}$
      • Amplitude $A = 3~\mathrm{m}$ (not needed for speed).

   2. Find the wave speed from $k$ and $\omega$

      The wave speed is $v = \frac{\omega}{k}$.

   $v = \frac{200}{4} = 50~\mathrm{m\,s^{-1}}.$

   3. Relate wave speed to tension and linear density For a string under tension $T$, the transverse wave speed is $v = \sqrt{\frac{T}{\mu}},$ …

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10. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
    Q.In a medium, a source produces 60 crests and 60 troughs in a time of 0.2 second. If the distance between a crest and its adjacent trough is 100 cm, then the speed of sound in the medium is (A) $600~\mathrm{ms}^{-1}$ (B) $1200~\mathrm{ms}^{-1}$ (C) $300~\mathrm{ms}^{-1}$ (D) $200~\mathrm{ms}^{-1}$

    ›Reveal solutionSolution

    The key is to find the frequency from the number of waves per second and the wavelength from the crest-to-trough distance. The speed comes out to $600~\mathrm{ms}^{-1}$, option (A).

    The problem gives you two separate pieces of information: one about how many waves pass per unit time, and one about the spatial size of a wave. The speed of a wave is always the product of its frequency and wavelength, $v = f \lambda$. So the task is to extract $f$ and $\lambda$ from the given data.

    Let’s break it down.

    1. Finding the frequency. The source produces 60 crests and 60 troughs in 0.2 seconds. One complete wave consists of one crest and one trough. So 60 crests and 60 troughs together mean 60 complete waves. The number of waves per second — the frequency — is therefore

    $f = \frac{60}{0.2} = 300~\text{Hz}.$

    2. Finding the wavelength. The distance between a crest and its adjacent trough is given as 100 cm. That distance is exactly half a wavelength (crest to trough is $\lambda/2$). So

    $\frac{\lambda}{2} = 100~\text{cm} = 1~\text{m},$

    which gives

    $$$$

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11. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.If the length of a string is $P$ when the tension in it is $6~\mathrm{N}$ and its length is $Q$ when the tension in it is $8~\mathrm{N}$, then the original length of the string is (A) $3P + 4Q$ (B) $3P - 4Q$ (C) $4P + 3Q$ (D) $4P - 3Q$

    ›Reveal solutionSolution

    The problem uses Hooke’s law for a string under tension: extension is proportional to tension. The original length is found by eliminating the spring constant from two equations, giving the result $4P - 3Q$.

    The key idea is that for an elastic string obeying Hooke’s law, the extension (change in length) is directly proportional to the applied tension, provided the string is not stretched beyond its elastic limit. So if the original (unstretched) length is $L$, then when tension is $T$, the length is $L + kT$, where $k$ is a constant (the reciprocal of the spring constant times the original length, but we don’t need that detail). We have two data points, and we can solve for $L$ without ever finding $k$.

    1. Let the original length of the string be $L$ (in metres, say). When the tension is $6~\mathrm{N}$, the length is $P$. So:

    $P = L + k \cdot 6$

    where $k$ is the constant of proportionality (extension per unit tension).

    2. Similarly, when the tension is $8~\mathrm{N}$, the length is $Q$:

    $Q = L + k \cdot 8$

    3. We have two linear equations in $L$ and $k$. To eliminate $k$, we can solve for $k$ from one equation and substitute into the other. From the first equation:

    $k = \frac{P - L}{6}$

    4. Substitute into the second equation:

    $Q = L + 8 \cdot \frac{P - L}{6}$

    5. Multiply both sides by 6 to clear the denominator:

    $6Q = 6L + 8(P - L)$

    6. Expand and simplify:

    $6Q = 6L + 8P - 8L$

    $$6Q = 8P - 2L …$$

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12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.A message signal of peak voltage 12 V is used to amplitude modulate a carrier signal of frequency 1.2 MHz. The amplitude of the side bands is (A) $12~\text{V}$ (B) $3~\text{V}$ (C) $6~\text{V}$ (D) $8~\text{V}$

    ›Reveal solutionSolution

    In standard AM, each sideband amplitude is half the modulating signal’s peak voltage. With a 12 V peak message, each sideband is 6 V, so the correct option is (C).

    The key idea here is that amplitude modulation (AM) creates two sidebands — one above and one below the carrier frequency — and each sideband carries half the amplitude of the modulating signal. This is not a guess; it follows directly from the mathematics of AM.

    Why this works:

    When you multiply a carrier wave $A_c \cos(\omega_c t)$ by a message signal $m(t) = A_m \cos(\omega_m t)$, the product expands using the cosine product identity into two new frequencies: $\omega_c + \omega_m$ and $\omega_c - \omega_m$. Each of these terms has amplitude $\frac{A_m}{2}$. That’s the sideband amplitude — no more, no less.

    Let’s walk through it step by step.

    1. Write the standard AM wave The general expression for an amplitude modulated wave is

    $s(t) = A_c \left[1 + \mu \cos(\omega_m t)\right] \cos(\omega_c t)$

    where $\mu = \frac{A_m}{A_c}$ is the modulation index. Here, the peak message voltage is given as 12 V, so $A_m = 12$ V. The carrier amplitude $A_c$ is not explicitly given, but that’s fine — we only need the sideband amplitude.

    2. Expand the product Using the identity $\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$, we get:

    $s(t) = A_c \cos(\omega_c t) + \frac{A_c \mu}{2} \cos\left[(\omega_c + \omega_m)t\right] + \frac{A_c \mu}{2} \cos\left[(\omega_c - \omega_m)t\right]$

    The first term is the carrier. The second and third terms are the upper and lower sidebands.

    3. Identify the sideband amplitude Each sideband has amplitude $\frac{A_c \mu}{2} = \frac{A_c \cdot (A_m / A_c)}{2} = \frac{A_m}{2}$ …

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