Physics

Simple musical instruments Questions

Chapter: Oscillations and Waves | Total Available: 29 Real Questions

Question 1 of 10 Authentic UPSC NDA Question

The time period of a 1 m long pendulum approximates to

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (c) is correct. The time period of a 1 m long pendulum approximates to 2 seconds.

πŸ“– Governing Law / Core Concept

The time period \( T \) of a simple pendulum is governed by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where:
  • T = Time period (seconds)
  • L = Length of the pendulum (meters)
  • g = Acceleration due to gravity (approximately 9.81 m/sΒ²)

πŸ§ͺ Step-by-Step Breakdown

In this case, let \( L = 1 \) m. Substitute \( L \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{1}{9.81}} \]
Calculate \( \sqrt{\frac{1}{9.81}} \):
\[ \sqrt{\frac{1}{9.81}} \approx 0.318 \]
Now, find \( T \):
\[ T \approx 2\pi \cdot 0.318 \approx 2.0 \text{ s} \]
Thus, the time period approximates to 2 seconds.

πŸ” Option Analysis

- Option (a) 6 s: Too long for a 1 m pendulum. - Option (b) 4 s: Exceeds expected range for this length. - Option (d) 1 s: Too short, inconsistent with calculations.

⚑ Mnemonic / Speed-Run

For estimating pendulum time periods, use \( T \approx 2\sqrt{L} \) when \( g \) is considered as approximately 10 m/sΒ² for quick mental calculations.

6. Visual Suggestion

6. Visual Suggestion

A diagram depicting a simple pendulum, showing the length \( L \), the pivot point, and the path of the bob swinging, aiding in visualizing the concept of the oscillation and its time period.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 15 (Oscillations).

Question 2 of 10 Authentic UPSC NDA Question

The length of a simple pendulum is increased four times to its previous value while the mass is doubled. What is the ratio of the new and previous time period of the pendulum?

Step-by-Step Logic & Solution

πŸ’‘ Solution

t:T 2:1

πŸ’‘ Hints & Visual Guide

πŸ“Œ Hints / Properties Used:

  • Time period of a simple pendulum is given as \( T = 2\pi \sqrt{\frac{l}{g}} \).
  • Mass of the bob does not affect the time period.

πŸ“Š Visual Diagram Suggestion:

[A labeled diagram of a simple pendulum indicating the length and its properties, including a side-by-side comparison of the time periods before and after the change in length and mass]

Question 3 of 10 Authentic UPSC NDA Question

The sound created in a big hall persists because of the repeated reflections. The phenomenon is called

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (a) is correct. The phenomenon is known as reverberation.

πŸ“– Governing Law / Core Concept

Reverberation occurs due to the repeated reflection of sound waves in large spaces. The basic principle is formulated as: \[ \text{Sound Intensity} \propto \frac{1}{d^2} \] where \( d \) is the distance from the sound source.

πŸ§ͺ Step-by-Step Breakdown

When sound waves encounter surfaces, they are reflected back multiple times before they dissipate. This results in the sound persisting in the environment. The reverberation time (\( T_r \)) can be calculated using Sabine's formula: \[ T_r = \frac{0.161 V}{A} \] where: - \( V \) = volume of the room (mΒ³) - \( A \) = total surface area of absorption (mΒ²) This reflection creates a layered effect of sound that can enhance auditory experiences, typical in concert halls.

πŸ” Option Analysis

- **B: Dispersion** refers to the separation of waves into their component frequencies, not applicable here. - **C: Refraction** is the bending of waves as they pass through different media, irrelevant to sound persistence. - **D: Diffraction** is the bending of waves around obstacles; it contributes to sound behavior but not to persistent sound in a hall.

⚑ Mnemonic / Speed-Run

To remember reverberation, think "Room Echo" β€” R for Room, E for Echo.

6. Visual Suggestion

6. Visual Suggestion

A diagram illustrating sound wave propagation and reflection within a hall, showing how sound reflects off walls creating multiple sound paths leading to a persistent auditory experience.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class 11 Physics, Chapter 15 (Waves).

Question 4 of 10 Authentic UPSC NDA Question

Which one of the following cannot be the unit of frequency of a sound wave?

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (a) is correct. dB (decibel) is not a unit of frequency.

πŸ“– Governing Law / Core Concept

The frequency of a sound wave is defined as the number of oscillations or cycles it completes per unit time. It is measured in Hertz (Hz), where \(1\, \text{Hz} = 1\, \text{s}^{-1}\).

πŸ§ͺ Step-by-Step Breakdown

The unit of frequency can be defined as:
\[ \text{Frequency} \, (f) = \frac{\text{Number of cycles}}{\text{Time interval}} \]
This leads us to the units: - Hertz (Hz) or cycles per second: \( \text{Hz} = \text{s}^{-1} \) - Minutes inverse (\( \text{min}^{-1} \)) can also be used, as it is simply cycles per minute. However, the decibel (dB) is a logarithmic unit used for ratios of power or intensity, not frequency.

πŸ” Option Analysis

- **A: dB** - Decibel, a unit for measuring sound intensity, not frequency. - **B: s⁻¹** - Equivalent to Hertz, valid unit of frequency. - **C: Hz** - Standard unit of frequency (1 Hz = 1 cycle/s). - **D: min⁻¹** - Valid unit representing frequency (cycles per minute).

⚑ Mnemonic / Speed-Run

Remember: "Hertz for cycles, seconds for time; anything else is off its prime."

6. Visual Suggestion

6. Visual Suggestion

Visualize a sound wave oscillating through time with its frequency labeled in Hz, dB, and min⁻¹. Illustrate a wave's upward and downward cycles within one second to relate distance between cycles to frequency.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 15 (Sound).

Question 5 of 10 Authentic UPSC NDA Question

Beats is a phenomenon that occurs when frequencies of two harmonic waves are

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (d) is correct. Beats occur due to two waves having nearly the same frequencies.

πŸ“– Governing Law / Core Concept

Beats are a phenomenon resulting from the interference of two harmonic waves with closely matched frequencies, represented mathematically by:

\[ f_{\text{beats}} = |f_1 - f_2| \]

where \( f_1 \) and \( f_2 \) are the frequencies of the two waves (in Hz). SI unit is Hertz (Hz).

πŸ§ͺ Step-by-Step Breakdown

When two waves \( y_1 = A \sin(2 \pi f_1 t) \) and \( y_2 = A \sin(2 \pi f_2 t) \) interfere, the resultant wave is given by:

\[ y = y_1 + y_2 = A \sin(2 \pi f_1 t) + A \sin(2 \pi f_2 t) \]

Using the sine addition formula, this can be simplified to observe beat frequencies. When \( f_1 \) and \( f_2 \) are nearly the same, the interference results in alternating constructive and destructive wave patterns, leading to beats.

πŸ” Option Analysis

Option (a): Equal frequencies lead to a steady wave, not beats. Option (b): Far apart frequencies produce no beats, only distinct waves. Option (c): Multiples lead to harmonics, not beats.

⚑ Mnemonic / Speed-Run

Remember: "Beats Occur when Frequencies are Close (BOFC)" to quickly identify the conditions for beat formation.

Visual Suggestion

Visual Suggestion

A wave diagram showing two waves with slightly different frequencies (e.g., 440 Hz and 442 Hz) illustrating how their superposition results in a beat frequency visible in amplitude variations.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 15 (Waves).

Question 6 of 10 Authentic UPSC NDA Question

Which one of the following statements is true for a simple harmonic oscillator?

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (b) is correct. The force acting on a simple harmonic oscillator is proportional to the displacement from mean position and directed opposite to it.

πŸ“– Governing Law / Core Concept

The motion of a simple harmonic oscillator is governed by Hooke's Law, which states that the restoring force \( F \) is proportional to the displacement \( x \) from equilibrium and acts in the opposite direction: \[ F = -kx \] where \( k \) is the spring constant (N/m).

πŸ§ͺ Step-by-Step Breakdown

The essential equations are: \[ a = \frac{F}{m} \] Given \( F = -kx \), substituting yields: \[ a = -\frac{k}{m}x \] Thus, acceleration is also proportional to displacement but opposite in direction, \( a \propto -x \).

πŸ” Option Analysis

- **Option A**: Incorrect, as the force is in the opposite direction of displacement. - **Option C**: Incorrect, acceleration varies with displacement; it is not constant. - **Option D**: Incorrect, the velocity of an oscillator in SHM is periodic.

⚑ Mnemonic / Speed-Run

Remember: "SHM is a dance: Force acts back!" (Force opposes displacement).

6. Visual Suggestion

6. Visual Suggestion

A schematic diagram illustrating the position of a mass on a spring, showing the equilibrium position, maximum displacement, and the opposite direction of the restoring force relative to the displacement.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter on Oscillations.

Question 7 of 10 Authentic UPSC NDA Question

Reverberation is a phenomenon associated with a

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option B is correct. Reverberation is caused by multiple reflections of sound.

πŸ“– Governing Law / Core Concept

Reverberation relates to the behavior of sound waves in enclosed spaces, governed by the principles of sound reflection.

πŸ§ͺ Step-by-Step Breakdown

Reverberation occurs when sound waves reflect off surfaces multiple times before they reach the listener's ears, extending the duration of sound perception. This is quantitatively described by the reverberation time, \( T_{60} \), defined as the time it takes for sound to decay by 60 dB.

\[ T_{60} = 0.161 \times \frac{V}{A} \]

Where:

  • V: Volume of the room (cubic meters, mΒ³)
  • A: Total surface area of the room (square meters, mΒ²)

This formula illustrates that reverberation time is shorter in rooms with more surface area (absorptive materials) and larger volumes.

πŸ” Option Analysis

Option A refers to refraction, which involves the bending of sound waves; Option C implies a single refraction, which doesn't apply; Option D mentions single reflection, inadequate for sound persistence. None of these describe reverberation.

⚑ Mnemonic / Speed-Run

Remember: "Reverberation Rings" - Multiple reflections create continuous sound.

6. Visual Suggestion

A schematic diagram showing sound waves bouncing off walls of a concert hall, visualizing how sound reflects multiple times, indicating increased reverberation time.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class 11 Physics, Chapter on Sound.

Question 8 of 10 Authentic UPSC NDA Question

Which among the following is true for propagation of sound waves?

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (b) is correct. Sound cannot travel in vacuum and it is a longitudinal wave in air.

πŸ“– Governing Law / Core Concept

Sound waves are mechanical waves governed by the principle that they require a medium (solid, liquid, or gas) to propagate. They are classified as longitudinal waves because the particle displacement is parallel to the wave direction.

πŸ§ͺ Step-by-Step Breakdown

1. Mechanical nature: Sound requires a medium, thus cannot travel through vacuum.

2. Wave type: In sound propagation, particles in the medium vibrate back and forth in the same direction as the wave travels, indicating a longitudinal wave motion.

Mathematically, the speed of sound in a gas can be expressed as:

\[ v = \sqrt{\frac{B}{\rho}} \]

where:

  • v = speed of sound (m/s)
  • B = bulk modulus of the medium (N/mΒ²)
  • \(\rho\) = density of the medium (kg/mΒ³)

πŸ” Option Analysis

  • A: Incorrect. Sound cannot travel in a vacuum.
  • C: Incorrect. Sound does not travel in vacuum.
  • D: Incorrect. Sound is not a transverse wave.

⚑ Mnemonic / Speed-Run

Remember: "Sound Needs a Medium" to recall that sound cannot travel through vacuum.

6. Visual Suggestion

6. Visual Suggestion

Illustrate a longitudinal wave with particle motion depicted in a spring or a slinky to show compression and rarefaction, highlighting that particle displacement is parallel to wave direction.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 15 (Waves).

Question 9 of 10 Authentic UPSC NDA Question

At 20° C, the speed of sound in water is approximately

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (c) is correct. The speed of sound in water at 20° C is approximately 1500 m/s.

πŸ“– Governing Law / Core Concept

The speed of sound in any medium is influenced by its density and elasticity. In water, the speed of sound is approximately given by the formula:

\[ v = \sqrt{\frac{E}{\rho}} \]

where \( v \) is the speed of sound, \( E \) is the modulus of elasticity, and \( \rho \) is the density of the medium.

πŸ§ͺ Step-by-Step Breakdown

At 20° C, the following values apply to water:

  • Density \( \rho \approx 1000 \, \text{kg/m}^3 \)
  • Bulk modulus (approximate elasticity) \( E \approx 2.2 \times 10^9 \, \text{N/m}^2 \)

Substituting these values into the formula:

\[ v = \sqrt{\frac{2.2 \times 10^9}{1000}} \]
\[ v = \sqrt{2.2 \times 10^6} \approx 1483 \, \text{m/s} \]

This approximation rounds to approximately 1500 m/s, affirming option (c).

πŸ” Option Analysis

  • Option (a) 330 m/s: Typical speed of sound in air at room temperature.
  • Option (b) 800 m/s: Incorrect; represents a low estimate.
  • Option (d) 5000 m/s: Exceeds normal values for any non-metallic medium.

⚑ Mnemonic / Speed-Run

Remember, the speed of sound in water is around 1500 m/s; think of it as swift as a swimmer!

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against standard physics texts and research on sound propagation in fluids.

Question 10 of 10 Authentic UPSC NDA Question

The loudness of sound depends upon the

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (b) is correct. Loudness is determined by the amplitude of sound waves.

πŸ“– Governing Law / Core Concept

The loudness of sound relates to its amplitude, represented in physics as:

\[ \text{Loudness} \propto A^2 \]

Where \( A \) is the amplitude of the sound wave.

πŸ§ͺ Step-by-Step Breakdown

Loudness is a subjective perception of sound intensity. It directly depends on the amplitude of the sound waves, which dictates how much energy is carried by the wave. Greater amplitude results in louder sounds.

πŸ” Option Analysis

  • A: Velocity of sound waves affects how sound travels but not its loudness.
  • C: Frequency affects pitch, not loudness.
  • D: Frequency and velocity together do not contribute to loudness; only amplitude does.

⚑ Mnemonic / Speed-Run

Remember: "Amplitude Affects Loudness!" This simple phrase helps recall that louder sounds come from greater wave amplitudes.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 12 (Sound).

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