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Waves appeared 22 times across 3 years — 2.5% of Physics. This question is from Wave Equation.

Year 2026 2025 2024 Total
Questions 7 10 5 22

The equation of a wave travelling on a string is y= [20π x+10π t] where x and t are distance and time in SI units. The minimum distance between two points having the same oscillating speed is : [cite: 120, 121]

Solution & Explanation

Related Formula
k = (2π)/(λ)

Δ xmin = (λ)/(2) [cite: 715]

Core Logic

From the given wave equation, the wave number k is the coefficient of x [cite: 120]:

k = 20π rad/m

Now find the wavelength λ: [cite: 717]

λ = (2π)/(k) = (2π)/(20π) = (1)/(10) m = 10 cm [cite: 717]

The minimum distance between any two points moving with identical speeds in a continuous wave cycle corresponds to a phase difference of π, which translates spatially to a half-wavelength separation ((λ)/(2)) [cite: 715]:

Distance = (λ)/(2) = (10)/(2) = 5 cm [cite: 717]

Pattern Recognition

Oscillating speed configuration reaches identical value fields twice per spatial wavelength cycle[cite: 715]. Thus, minimum distance separation scales exactly to (λ)/(2)[cite: 715].

Chapter Mix

Class 11 Physics: Waves

Reference Study Guides

More Waves Previous-Year Questions — Page 5

Q37 jee_main_2024_31_jan_evening Speed of Sound in Gases
The speed of sound in oxygen at S.T.P. will be approximately: (Given, R = 8.3 J K⁻¹ , γ = 1.4)
  • A. 310 m/s
  • B. 333 m/s
  • C. 341 m/s
  • D. 325 m/s

Solution

Related Formula
v = √((γ RT)/(M))
Core Logic

For Oxygen (O₂) at standard temperature and pressure (S.T.P.): T = 273 K M = 32 g/mol = 32 × 10⁻³ kg/mol γ = 1.4 R = 8.3 J K⁻¹ mol⁻¹

Step 1: Calculate Velocity
v = 1.4 × 8.3 × 27332 × 10⁻³ v = 3172.2632 × 10⁻³ v = √(99.133 × 10³) v = √(99133) ≈ 314.85 m/s

Approximating to the closest given option yields 310 m/s.

Pattern Recognition

For diatomic gases around room temp or STP, velocities range roughly from 250 to 350 m/s depending on molar mass (N₂ ≈ 334, O₂ ≈ 315). Recognize 315 is closest to option (1) due to standard approximations taken in exams.

Chapter Mix

Class 11 Physics: Waves Class 11 Physics: Kinetic Theory of Gases

Q jee_main_2024_31_jan_morning Organ Pipes
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60~cm, the length of the closed pipe will be:
  • A. 60~cm
  • B. 45~cm
  • C. 30~cm
  • D. 15~cm

Solution

Related Formula
fclosed, fundamental = (v)/(4Lc) fopen, 1st overtone = (2v)/(2Lₒ)
Core Logic

Organ Pipes diagram for Q45 - JEE Main 2024 Morning
Organ Pipes diagram for Q45 - JEE Main 2024 Morning

Organ Pipes diagram for Q45 - JEE Main 2024 Morning
Organ Pipes diagram for Q45 - JEE Main 2024 Morning

For a closed organ pipe, the fundamental frequency (1st harmonic) is:

f₁ = (v)/(λ) = (v)/(4L₁)

where L₁ is the length of the closed pipe.

For an open organ pipe, the first overtone (2nd harmonic) is:

f₂ = (2v)/(2L₂) = (v)/(L₂)

where L₂ is the length of the open pipe (L₂ = 60 cm).

Step 2: Equating Frequencies

Given f₁ = f₂:

(v)/(4L₁) = (v)/(L₂)

L₂ = 4L₁

60 = 4 × L₁ L₁ = 15 cm
Chapter Mix

Class 11 Physics: Waves

More Waves Questions — jee_main_2025_07_april_evening

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