Disaster Management · TEE · practice

Magnitude comparisons — optional authored practice

Optional calculation practice

Supplementary worked examples · no numerical question found in the available 12 TEE sittings

Source of the calculations. M2 PDF p 55 teaches the tenfold amplitude relation. The numerical values below were added for explanation; none is a recovered PYQ. The approximate energy-ratio relation is an external scientific extension (USGS), not a formula explicitly found in the supplied lecture text.

Definition. A logarithmic scale expresses multiplicative changes as differences. A base-10 logarithm asks: “10 raised to what power gives this number?”

Simple meaning. Log bas ulta sawaal hai: 10 ki kaunsi power?

For example, 10² = 100, so log₁₀(100) = 2. On a compatible amplitude-based magnitude scale, a difference of 1 corresponds to a factor of 10 in measured amplitude after the required corrections. Earthquake energy follows the approximate relationship below; energy and amplitude are different quantities.

\Delta M=M_2-M_1,\qquad \frac{A_2}{A_1}=10^{\Delta M}LaTeX: \Delta M=M_2-M_1,\qquad \frac{A_2}{A_1}=10^{\Delta M}
\frac{E_2}{E_1}\approx10^{1.5\Delta M}LaTeX: \frac{E_2}{E_1}\approx10^{1.5\Delta M}

A is wave amplitude; E is released energy; Δ means “change in”. Amplitude comparisons assume the relevant scale and comparable/corrected recording conditions. Do not turn an energy ratio into a claim about building damage.

Illustrative calculation (not a PYQ). Compare the amplitude and approximate energy of earthquakes of magnitude 6 and 5.

Answer

Given

M₂ = 6; M₁ = 5.

To find

A₂/A₁ and E₂/E₁.

Formula and why

A_2/A_1=10^{\Delta M};\quad E_2/E_1\approx10^{1.5\Delta M}LaTeX: A_2/A_1=10^{\Delta M};\quad E_2/E_1\approx10^{1.5\Delta M}

The magnitude difference is the exponent in the respective logarithmic relationship.

Step-by-step solution

Step 1. Find the difference

\Delta M=6-5=1LaTeX: \Delta M=6-5=1

Step 2. Find amplitude ratio

A_2/A_1=10^1=10LaTeX: A_2/A_1=10^1=10

Step 3. Find energy ratio

E_2/E_1\approx10^{1.5}=31.6228LaTeX: E_2/E_1\approx10^{1.5}=31.6228

Final answer: 10 times the amplitude and approximately 31.6 times the energy.

Illustrative calculation (not a PYQ). Compare magnitude 7 with magnitude 5 using amplitude and energy ratios.

Answer

Given

M₂ = 7; M₁ = 5.

To find

The two ratios.

Formula and why

A_2/A_1=10^{\Delta M};\quad E_2/E_1\approx10^{1.5\Delta M}LaTeX: A_2/A_1=10^{\Delta M};\quad E_2/E_1\approx10^{1.5\Delta M}

There are two magnitude units between the events, so multiply factors through the exponent.

Step-by-step solution

Step 1. Subtract magnitudes

\Delta M=7-5=2LaTeX: \Delta M=7-5=2

Step 2. Calculate amplitude

10^2=100LaTeX: 10^2=100

Step 3. Calculate energy

10^{1.5\times2}=10^3=1000LaTeX: 10^{1.5\times2}=10^3=1000

Final answer: 100 times the amplitude; approximately 1,000 times the energy.

Illustrative calculation (not a PYQ). A corrected amplitude ratio is 100. Find the magnitude difference. If the smaller event has magnitude 4.3, estimate the larger event’s magnitude.

Answer

Given

A₂/A₁ = 100; M₁ = 4.3.

To find

ΔM and M₂.

Formula and why

\Delta M=\log_{10}(A_2/A_1),\quad M_2=M_1+\Delta MLaTeX: \Delta M=\log_{10}(A_2/A_1),\quad M_2=M_1+\Delta M

The logarithm reverses exponentiation: 10² is 100.

Step-by-step solution

Step 1. Take the base-10 logarithm

\Delta M=\log_{10}(100)=2LaTeX: \Delta M=\log_{10}(100)=2

Step 2. Add the difference

M_2=4.3+2=6.3LaTeX: M_2=4.3+2=6.3

Final answer: Magnitude difference 2; larger magnitude 6.3.

Calculator: use the log key for base 10, not ln. For 101.5, use the calculator’s 10ˣ function with exponent 1.5. Keep brackets around a multiplied exponent.

Yaad rakho: +1 magnitude: ×10 amplitude, about ×32 energy.