Daily Olympiad: Physics - Oscillations Waves [20260511]

Challenge yourself with today's JEE Main practice! This test covers 'Oscillations Waves' for Physics (JEE Main - 12). Level: Hard | Duration: 45 mins.

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1. A particle executes SHM with a time period T. If its maximum velocity is v_max and maximum acceleration is a_max, what is the ratio of their amplitudes of acceleration to velocity?

Solution
Correct: A
In SHM, the maximum velocity is v_max = Aω and maximum acceleration is a_max = Aω². The ratio a_max/v_max = ω = 2π/T. Since ω = √(a_max/v_max²), the ratio of amplitudes (Aω/Aω²) gives 1/ω. Hence, the ratio is a_max / v_max.

2. Two wave functions y₁ = 4sin(2x - t) and y₂ = 4sin(2x + t) superpose. What is the amplitude of the resultant standing wave at x = λ/4?

Solution
Correct: B
Superposition gives y = 8cos(2x)sin(t). At x = λ/4, cos(2x) = cos(π) = -1. Amplitude is |8cos(2x)| = 8. Hence, the maximum amplitude is 8.

3. A closed pipe resonates with a tuning fork at its fundamental frequency. If the open end is suddenly converted to an open pipe, the new fundamental frequency becomes

Solution
Correct: B
Fundamental frequency of closed pipe is f = v/(4L). For open pipe, f' = v/(2L). Hence, f' = 2f.

4. A wave pulse on a string reflects off a rigid boundary. The reflected pulse has

Solution
Correct: B
Reflection from a rigid boundary inverts the wave due to Newton's third law (equal and opposite force). Hence, the phase changes by π.

5. A spring-mass system oscillates horizontally with amplitude A. When it reaches x = A/2, its potential energy is E_p. What is its kinetic energy at x = -A/2?

Solution
Correct: A
Potential energy in SHM is E_p = (1/2)kx². At x = ±A/2, potential energy is (1/2)k(A²/4) = E_p. Kinetic energy is total energy (KE + PE). Since energy is conserved, KE at x = ±A/2 is same as E_p.

6. A tuning fork of frequency 256 Hz is moving towards a wall at 5 m/s. The beat frequency heard by an observer moving away from the wall at 1 m/s is (speed of sound = 340 m/s)

Solution
Correct: C
Apparent frequency from source approaching wall: f' = (340/(340 - 5)) * 256 = 259.1 Hz. Reflected sound: observer moves away, f'' = ((340 - 1)/340) * 259.1 = 256.4 Hz. Beat frequency is |259.1 - 256.4| ≈ 2.7 Hz ≈ 3.6 Hz.

7. Two identical pendulums are coupled with a weak spring. The system undergoes normal mode oscillations with frequencies f₁ and f₂. If their individual frequencies are f₀, what is the relation between f₁ and f₂?

Solution
Correct: A
In coupled pendulums, the normal mode frequencies satisfy f₁² + f₂² = 2f₀². When coupling is weak, f₁ ≈ f₀ - Δ and f₂ ≈ f₀ + Δ, leading to f₁f₂ ≈ f₀².

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