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#amplitude calculation

3 public questions tagged with this topic.

Superposition, Resultant Amplitude and Intensity

What is the resultant amplitude of two coherent waves of amplitude \( a \) with a phase difference of \( 7\pi/2 \)?

**Superposition principle** resultant displacement sum of individual, for two coherent waves amplitude a each, resultant amplitude A = √(a² + a² +2a² cosφ)=2a|cos(φ/2)|, phase difference φ, path difference Δ = (φ/2π)λ, for φ=π/2 A=√2 a, for φ=6π cos3π=-1? Actually φ=6π cos3π? A=2a|cos3π|=2a, for φ=4π A=2a, intensity I ∝ A², maximum I_max=4I₀ when φ=0, I=2I₀(1+cosφ)=4I₀ cos²(φ/2). Resultant amplitude A = 2a cos(Φ/2) . For Φ = (7π/2) , A = 2a cos((7π/4)) = 2a ((√(2)/2)) = a√(2) . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n

Ref: NCERT > Physics Book > Wave Optics > Superposition, Resultant Amplitude and Intensity

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Energy in SHM - Kinetic, Potential and Total

A spring of \( k = 360 \, \text{N/m} \) has a \( 1.5 \, \text{kg} \) mass. If \( E = 1.8 \, \text{J} \), what is the amp

**Energy in SHM** interconverts between kinetic K = ½ m v² = ½ m ω² (A² - x²) and potential U = ½ k x² = ½ m ω² x², total E = K + U = ½ k A² = ½ m ω² A² constant, independent of time. At mean position x=0, E = K_max = ½ m ω² A², at extremes x=±A, E = U_max = ½ k A². Total energy: E = (1/2) k A² . 1.8 = 0.5 × 360 × A² ⇒ 1.8 = 180 A² ⇒ A² = 0.01 ⇒ A = 0.1 m . Applying x = A

Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total

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Superposition and Interference of Waves

Two waves \( y_1 = 3 \sin (4x - 10t) \) and \( y_2 = 3 \sin (4x - 10t + \pi/2) \) interfere. What is the amplitude of th

**Superposition principle** states resultant displacement equals algebraic sum of individual waves, y = y₁ + y₂. For coherent waves with phase difference φ, resultant amplitude A = √(a₁² + a₂² + 2a₁a₂ cosφ), equal amplitudes give A = 2a cos(φ/2), constructive when φ = 2nπ, destructive when φ = (2n+1)π. Amplitude: A = 2a cos (Φ/2) , where a = 3 m , Φ = π/2 . A = 2 × 3 cos π/4 = 6 × (1/√(2)) = 3√(2) ≈ 4.24 m . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 4.2 m, illustrating frequency-length-speed interdependence and quantization by boundaries

Ref: NCERT > Physics Book > Waves > Superposition and Interference of Waves

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