Chapter Exam

Unit 10: Oscillations and Waves — Online MCQ Test

Q: A closed organ pipe and an open organ pipe are filled by two different gases having same bulk modulus but different densities ρ1 and ρ2, respectively. The frequency of 9th harmonic of closed tube is identical with 4th harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is ρ1:ρ2 = 1:16, then the length of the open tube is:

Answer: D. 20/9 cm — Frequency f = v/λ, where v = √(B/ρ). For a closed pipe f_closed = n1*v1/(4L1) and for an open pipe f_open = n2*v2/(2L2). Setting these equal and substituting the density ratios allows solving for L2.

Q: The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s. The frequency of this simple harmonic oscillator is Hz. (take π = 22/7)

Answer: A. 14 — The kinetic energy KE = 1/2 mv^2 = 1/2 m(Aω cos(ωt))^2 = 1/2 mA^2ω^2 * cos^2(ωt). The angular frequency of KE oscillation (ω_KE) is 2ω_osc. Given ω_KE = 176 rad/s, ω_osc = 88 rad/s. Frequency f = ω_osc / (2π) = 88 / (2 * 22/7) = 88 * 7 / 44 = 14 Hz.

Q: The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s. The frequency of this simple harmonic oscillator is Hz. (take π = 22/7)

Answer: A. 14 — The kinetic energy of an SHO oscillates at twice the frequency of the oscillator. Given ω_KE = 176 rad/s, the frequency of KE oscillation is f_KE = ω_KE / (2π) = 176 / (2 * 22/7) = 176 * 7 / 44 = 28 Hz. Since f_KE = 2 * f_oscillator, then f_oscillator = 28 / 2 = 14 Hz.

Q: Using a simple pendulum experiment g is determined by measuring its time period T. Which of the following plots represent the correct relation between the pendulum length L and time period T?

Answer: B. T^2 vs L (Linear) — The time period of a simple pendulum is T = 2π√(L/g). Squaring both sides, T^2 = (4π^2/g) * L. Thus, the plot of T^2 vs L is a straight line passing through the origin.

Q: A carrier wave of amplitude $15\ \text{V}$ is modulated by a sinusoidal base band signal of amplitude $3\ \text{V}$. The ratio of maximum amplitude to minimum amplitude in an amplitude modulated wave is

Answer: D. $\frac{3}{2}$ — For amplitude modulation, \[ A_{\max} = A_c + A_m, \qquad A_{\min} = A_c - A_m. \] Here $A_c = 15\ \text{V}$ and $A_m = 3\ \text{V}$. So \[ A_{\max} = 18, \qquad A_{\min} = 12. \] Therefore \[ \frac{A_{\max}}{A_{\min}} = \frac{18}{12} = \frac{3}{2}. \] Hence the correct answer is $\frac{3}{2}$.

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Practice Unit 10: Oscillations and Waves with a free chapter-wise online MCQ test for JEE JEE MAINS PHYSICS. This chapter covers: Oscillations and periodic motion: time period frequency displacement as a function of time periodic functions. Simple harmonic motion (S.H.M.) and its equation phase oscillations o.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Unit 10: Oscillations and Waves — Important Questions & Answers (FAQ)

Frequently asked questions from JEE JEE MAINS PHYSICS — Unit 10: Oscillations and Waves, with answers and explanations. These are sample questions; the exam has its own separate question set.

A closed organ pipe and an open organ pipe are filled by two different gases having same bulk modulus but different densities ρ1 and ρ2, respectively. The frequency of 9th harmonic of closed tube is identical with 4th harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is ρ1:ρ2 = 1:16, then the length of the open tube is:

A. 15/7 cm
B. 20/7 cm
C. 15/9 cm
D. 20/9 cm ✓

Answer: D. 20/9 cm
Frequency f = v/λ, where v = √(B/ρ). For a closed pipe f_closed = n1*v1/(4L1) and for an open pipe f_open = n2*v2/(2L2). Setting these equal and substituting the density ratios allows solving for L2.

The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s. The frequency of this simple harmonic oscillator is Hz. (take π = 22/7)

A. 14 ✓
B. 88
C. 28
D. 176

Answer: A. 14
The kinetic energy KE = 1/2 mv^2 = 1/2 m(Aω cos(ωt))^2 = 1/2 mA^2ω^2 * cos^2(ωt). The angular frequency of KE oscillation (ω_KE) is 2ω_osc. Given ω_KE = 176 rad/s, ω_osc = 88 rad/s. Frequency f = ω_osc / (2π) = 88 / (2 * 22/7) = 88 * 7 / 44 = 14 Hz.

The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s. The frequency of this simple harmonic oscillator is Hz. (take π = 22/7)

A. 14 ✓
B. 88
C. 28
D. 176

Answer: A. 14
The kinetic energy of an SHO oscillates at twice the frequency of the oscillator. Given ω_KE = 176 rad/s, the frequency of KE oscillation is f_KE = ω_KE / (2π) = 176 / (2 * 22/7) = 176 * 7 / 44 = 28 Hz. Since f_KE = 2 * f_oscillator, then f_oscillator = 28 / 2 = 14 Hz.

Using a simple pendulum experiment g is determined by measuring its time period T. Which of the following plots represent the correct relation between the pendulum length L and time period T?

A. T vs L (Linear)
B. T^2 vs L (Linear) ✓
C. L vs T^2 (Inverse)
D. T vs L^2 (Linear)

Answer: B. T^2 vs L (Linear)
The time period of a simple pendulum is T = 2π√(L/g). Squaring both sides, T^2 = (4π^2/g) * L. Thus, the plot of T^2 vs L is a straight line passing through the origin.

A carrier wave of amplitude $15\ \text{V}$ is modulated by a sinusoidal base band signal of amplitude $3\ \text{V}$. The ratio of maximum amplitude to minimum amplitude in an amplitude modulated wave is

A. 2
B. 5
C. 1
D. $\frac{3}{2}$ ✓

Answer: D. $\frac{3}{2}$
For amplitude modulation, \[ A_{\max} = A_c + A_m, \qquad A_{\min} = A_c - A_m. \] Here $A_c = 15\ \text{V}$ and $A_m = 3\ \text{V}$. So \[ A_{\max} = 18, \qquad A_{\min} = 12. \] Therefore \[ \frac{A_{\max}}{A_{\min}} = \frac{18}{12} = \frac{3}{2}. \] Hence the correct answer is $\frac{3}{2}$.

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