BPHE-102

BPHE-102: Oscillations and Waves – All Assignment Answers (2025)

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BPHE-102: Oscillations and Waves – All Questions Answered Below are the complete answers to the IGNOU assignment for BPHE-102: Oscillations and Waves (2025). Click on each question to view its detailed solution. a) A simple harmonic motion is represented by x(t) = a cos(ωt). Obtain expressions for velocity and acceleration of the oscillator. Also, plot […]

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e) Consider two cylindrical pipes of equal length. One of these acts as a closed organ pipe and the other as open organ pipe. The frequency of the third harmonic in the closed pipe is 200 Hz higher than the first harmonic of the open pipe. Calculate the fundamental frequency of the closed pipe.

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Introduction This problem involves harmonics in open and closed cylindrical organ pipes of equal length. We are given a relation between harmonics and are required to find the fundamental frequency of the closed pipe. Understanding the Harmonics In an open organ pipe, all harmonics are present: f₁, 2f₁, 3f₁, … In a closed organ pipe,

e) Consider two cylindrical pipes of equal length. One of these acts as a closed organ pipe and the other as open organ pipe. The frequency of the third harmonic in the closed pipe is 200 Hz higher than the first harmonic of the open pipe. Calculate the fundamental frequency of the closed pipe. Read More »

d) A stretched string of mass 20 g vibrates with a frequency of 30 Hz in its fundamental mode and the supports are 40 cm apart. The amplitude of vibrations at the antinode is 4 cm. Calculate the velocity of propagation of the wave in the string as well as the tension in it.

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Introduction This question involves calculating the wave velocity and tension in a stretched string vibrating in its fundamental mode. We are given frequency, length, and mass of the string, and are asked to calculate the wave speed and tension. Given Mass (m) = 20 g = 0.020 kg Length (L) = 40 cm = 0.40

d) A stretched string of mass 20 g vibrates with a frequency of 30 Hz in its fundamental mode and the supports are 40 cm apart. The amplitude of vibrations at the antinode is 4 cm. Calculate the velocity of propagation of the wave in the string as well as the tension in it. Read More »

c) The linear density of a vibrating string is 1.3 × 10⁻⁴ kg/m. A transverse wave is propagating on the string and is described by the equation y(x, t) = 0.021 sin(30t − x), where x and y are in metres and t is in seconds. Calculate the tension in the string.

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Introduction We are given the wave function of a transverse wave traveling along a string. The wave equation is: y(x, t) = 0.021 sin(30t − x) We are also given: Linear mass density (μ) = 1.3 × 10⁻⁴ kg/m The goal is to determine the tension (T) in the string. Step 1: Identify Wave Parameters

c) The linear density of a vibrating string is 1.3 × 10⁻⁴ kg/m. A transverse wave is propagating on the string and is described by the equation y(x, t) = 0.021 sin(30t − x), where x and y are in metres and t is in seconds. Calculate the tension in the string. Read More »

b) A sinusoidal wave is described by y(x, t) = 3.0 sin(3.52t − 2.01x) cm. Determine the amplitude, wave number, wavelength, frequency and velocity of the wave.

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Introduction In this problem, we are given the equation of a sinusoidal wave in the form: y(x, t) = 3.0 sin(3.52t − 2.01x) (in cm) This standard wave equation allows us to extract different parameters of wave motion like amplitude, wave number, wavelength, angular frequency, and wave speed. Step-by-Step Extraction of Parameters 1. Amplitude (A)

b) A sinusoidal wave is described by y(x, t) = 3.0 sin(3.52t − 2.01x) cm. Determine the amplitude, wave number, wavelength, frequency and velocity of the wave. Read More »

a) The oscillations of two points x₁ and x₂ at x = 0 and x = 1 m respectively are modelled as: y₁ = 0.3 sin(4πt) and y₂ = 0.3 sin(4πt + π/8). Calculate the wavelength and speed of the associated wave.

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Introduction In this problem, we are given the oscillations at two different points on a wave. By analyzing the phase difference and distance between these points, we can determine the wavelength and speed of the wave. This is a typical problem related to wave propagation and phase relationships. Given y₁ = 0.3 sin(4πt) at x

a) The oscillations of two points x₁ and x₂ at x = 0 and x = 1 m respectively are modelled as: y₁ = 0.3 sin(4πt) and y₂ = 0.3 sin(4πt + π/8). Calculate the wavelength and speed of the associated wave. Read More »

e) Establish the equation of motion of a weakly damped forced oscillator explaining the significance of each term. Differentiate between transient and steady state of the oscillator.

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Introduction A forced oscillator is one that is subjected to a periodic external force. When damping is small, it is called a weakly damped forced oscillator. This system shows interesting behavior such as resonance and steady-state oscillation. In this question, we derive the equation of motion and interpret its terms. We also distinguish between the

e) Establish the equation of motion of a weakly damped forced oscillator explaining the significance of each term. Differentiate between transient and steady state of the oscillator. Read More »

d) For a damped harmonic oscillator, the equation of motion is m d²x/dt² + γ dx/dt + kx = 0 with m = 0.50 kg, γ = 0.70 kg/s and k = 70 N/m. Calculate (i) the period of motion, (ii) number of oscillations in which its amplitude will become half of its initial value, (iii) the number of oscillations in which its mechanical energy will drop to half of its initial value, (iv) its relaxation time, and (v) quality factor.

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Introduction This is a damped harmonic oscillator problem where we need to calculate multiple physical parameters based on the values of mass, damping constant, and spring constant. The motion is governed by the differential equation: m d²x/dt² + γ dx/dt + kx = 0 Given: m = 0.50 kg γ = 0.70 kg/s k =

d) For a damped harmonic oscillator, the equation of motion is m d²x/dt² + γ dx/dt + kx = 0 with m = 0.50 kg, γ = 0.70 kg/s and k = 70 N/m. Calculate (i) the period of motion, (ii) number of oscillations in which its amplitude will become half of its initial value, (iii) the number of oscillations in which its mechanical energy will drop to half of its initial value, (iv) its relaxation time, and (v) quality factor. Read More »

c) Two collinear harmonic oscillations x₁ = 8 sin (100πt) and x₂ = 12 sin (96πt) are superposed. Calculate the values of time when the amplitude of the resultant oscillation will be (i) maximum and (ii) minimum.

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Introduction We are given two harmonic oscillations: x₁ = 8 sin(100πt) x₂ = 12 sin(96πt) These are sinusoidal functions with different frequencies. When superposed, they result in a phenomenon known as beats. The amplitude of the resultant wave varies with time due to interference between the two waves. Step 1: Beat Frequency Let the two

c) Two collinear harmonic oscillations x₁ = 8 sin (100πt) and x₂ = 12 sin (96πt) are superposed. Calculate the values of time when the amplitude of the resultant oscillation will be (i) maximum and (ii) minimum. Read More »

b) The time period of a simple pendulum, called ‘seconds pendulum’, is 2 s. Calculate the length, angular frequency and frequency of the pendulum. What is the difference between a simple pendulum and a compound pendulum?

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Introduction A seconds pendulum is defined as a simple pendulum whose time period is exactly 2 seconds. In this question, we calculate its length, angular frequency, and frequency. Additionally, we differentiate between a simple and compound pendulum. Given Time period (T) = 2 s Acceleration due to gravity (g) = 9.8 m/s² Step 1: Calculate

b) The time period of a simple pendulum, called ‘seconds pendulum’, is 2 s. Calculate the length, angular frequency and frequency of the pendulum. What is the difference between a simple pendulum and a compound pendulum? Read More »

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