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Wave Attenuation Calculator

Find intensity after traveling distance x in an absorbing medium

Parameters

W/m²ⓘ
m⁻¹ⓘ
mⓘ
Show Trail

Controls

xⓘ

Calculated Values

Final Intensity:
0.22;W/m20.22;W/m²
Level Drop:
−6.51;dB-6.51;dB

Examples

Moderate absorption

α = 0.5 m⁻¹, x = 3 m.

    Shallow depth

    α = 1 m⁻¹, x = 1 m.

      Visualization

      Exponential Attenuation of Waves in Absorbing Media

      As waves propagate through a medium that absorbs or scatters energy, the intensity (or amplitude) decreases with distance. For many materials, the decrease is approximately exponential: I(x) = I₀ e^(−αx), where I₀ is intensity at x = 0 and α is the attenuation coefficient (m⁻¹).

      Larger α means faster decay. α depends on material, frequency, temperature, and humidity. Ultrasound in tissue has much larger α at 5 MHz than at 1 MHz — higher frequencies give better resolution but penetrate less.

      Half-intensity depth (penetration depth) x₁/₂ = ln(2)/α ≈ 0.693/α is the distance at which I drops to half I₀. After distance 2x₁/₂, intensity is one-quarter; after nx₁/₂, I = I₀/2ⁿ.

      In decibels, attenuation over distance x is Δβ = 10 log(I/I₀) = −10 log(e) × αx ≈ −4.34 αx dB when α is in m⁻¹. Engineers often tabulate dB/m for cables, seawater, and concrete.

      Exponential attenuation (absorption) is different from geometric spreading of a point source (I ∝ 1/r²). Real situations often have both: I = (P/(4πr²)) e^(−αr).

      Example: I₀ = 1 W/m², α = 0.5 m⁻¹, x = 3 m → I = e^(−1.5) ≈ 0.223 W/m², a drop of about 6.5 dB. Medical imaging must balance α, frequency, and power limits.

      Class 12 may mention damping of oscillations; exponential decay is the spatial analog. JEE can ask x₁/₂, percent transmitted, or compare two materials.

      Key Concepts

      • I = I₀ e^(−αx)
      • α = attenuation coefficient (m⁻¹)
      • x₁/₂ = ln(2)/α
      • Δβ ≈ −4.34 αx dB
      • Absorption vs 1/r² spreading
      • α increases with f in many media

      Real-World Applications

      • Medical ultrasound penetration limits
      • Sonar range in seawater
      • Fiber optic and RF cable loss (dB/km)
      • Acoustic barriers and anechoic materials
      • Class 12–JEE attenuation numericals

      Explore Further

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      • Wave Reflection

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      • Wave Transmission

        Analyze wave transmission between different media and wavelength changes.

      Physics Equations

      Attenuation:
      I=I0e−αxI = I_0 e^{-\alpha x}

      Step-by-Step Solution

      See how the main results are calculated.

      1

      Step 1: Exponential Attenuation

      Equation:

      I=I0e−αxI = I_0 e^{-\alpha x}

      Explanation:

      α = attenuation coefficient (m⁻¹); x = distance through medium.

      2

      Step 2: Given

      Result:

      I0=1,α=0.5m−1,x=3mI₀ = 1, α = 0.5 m⁻¹, x = 3 m
      3

      Step 3: Exponent

      Calculation:

      −αx=−0.5×3=−1.5000-\alpha x = -0.5 \times 3 = -1.5000
      4

      Step 4: Intensity

      Calculation:

      I=1×e−1.5000=0.223130I = 1 \times e^{-1.5000} = 0.223130

      Result:

      I=0.223130I = 0.223130
      5

      Step 5: Decibel Drop

      Calculation:

      Δβ=10log⁡(I/I0)=−6.51dB\Delta\beta = 10\log(I/I_0) = -6.51 dB

      Explanation:

      Negative value means attenuation.

      6

      Step 6: Half-Intensity Distance

      Equation:

      x1/2=ln⁡2αx_{1/2} = \frac{\ln 2}{\alpha}

      Calculation:

      x1/2=1.3863mx_{1/2} = 1.3863 m

      Explanation:

      Distance for intensity to drop by half.

      Frequently Asked Questions (FAQ)

      Attenuation vs absorption?

      Absorption converts wave energy to heat. Attenuation includes absorption plus scattering out of the beam direction.

      Does amplitude decay the same way?

      For linear waves, amplitude A ∝ √I, so A(x) = A₀ e^(−αx/2) — half the exponent of intensity.

      Can α be negative?

      Not for passive media. Amplification (negative loss) requires an active energy source.

      Percent transmitted?

      I/I₀ = e^(−αx) × 100%. At x = x₁/₂, 50% transmitted.

      Why higher frequency ultrasound penetrates less?

      α generally increases with frequency in soft tissue due to absorption mechanisms.

      Practice MCQs

      1. Attenuation law:
      2. Larger α means:
      3. x₁/₂ =
      4. Double distance x:
      5. α units:
      6. At x=0:

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