The speed $ v $ of a wave on a string depends on the tension $ F $ in the string and the mass per unit length $ m/L $ of the string. If it is known that [F] = [ML][T]–2, the values of the constants $ a $ and $ b $ in the following equation for the speed of a wave on a string are: \[ v = (\text{constant}) F^a \left( \frac{m}{L} \right)^b \]

Question

A ball is thrown vertically upwards with an initial velocity of $ 20 \, \text{m/s} $. How high will the ball rise? (Take $ g = 10 \, \text{m/s}^2 $)

Answer 1

Step 1: Recall the physical relationship.
For a stretched string, the wave speed is known to be $ v = \sqrt{\dfrac{F}{m/L}} $, which can be written as $ v = F^{1/2} \left( \dfrac{m}{L} \right)^{-1/2} $.

Step 2: Check the dimensions.
With $ [F] = [M][L][T]^{-2} $ and $ [m/L] = [M][L]^{-1} $, raising the first to the power $ \frac{1}{2} $ and the second to the power $ -\frac{1}{2} $ gives $ [M]^{0}[L]^{1}[T]^{-1} $, which is exactly the dimension of speed.

Step 3: Read off the exponents.
Comparing with $ v = (\text{constant}) F^a (m/L)^b $, this gives $ a = \dfrac{1}{2} $ and $ b = -\dfrac{1}{2} $.
\[ \boxed{a = \tfrac{1}{2}, \; b = -\tfrac{1}{2} \; (\text{option 1})} \]

Answer 2

Instead of working with abstract dimension symbols, we can plug in the actual SI units, kilograms, metres and seconds, for each quantity and require the two sides of the equation to carry the same units.

$ a = \frac{1}{2}, b = \frac{1}{2} $: This is the combination that keeps the square-root treatment consistent on both the tension and mass-per-length sides.
$ a = 2, b = -1 $: Squaring the tension unit and inverting the mass-per-length unit leaves $ \text{kg}^{1} $ uncancelled, so the units do not reduce to $ \text{m}\cdot\text{s}^{-1} $ here.
$ a = \frac{1}{2}, b = -1 $: This leaves $ \text{kg}^{-1/2} $ unaccounted for, so the units again fail to reduce to plain metres per second.
$ a = 1, b = \frac{1}{2} $: This leaves $ \text{kg}^{3/2} $ uncancelled, the largest mismatch of the four, so it is not consistent either.

Working purely in kilograms, metres and seconds, the only way every kilogram cancels and the metres and seconds reduce to $ \text{m}\cdot\text{s}^{-1} $ is with $ a = \dfrac{1}{2} $ and $ b = -\dfrac{1}{2} $, the square-root-of-tension over square-root-of-mass-per-length combination described by the first option.

So the correct answer is $ a = \frac{1}{2}, b = -\frac{1}{2} $, the first option.

Answer 3

Instead of working with abstract dimension symbols, we can plug in the actual SI units, kilograms, metres and seconds, for each quantity and require the two sides of the equation to carry the same units.

$ a = \frac{1}{2}, b = \frac{1}{2} $: This is the combination that keeps the square-root treatment consistent on both the tension and mass-per-length sides.
$ a = 2, b = -1 $: Squaring the tension unit and inverting the mass-per-length unit leaves $ \text{kg}^{1} $ uncancelled, so the units do not reduce to $ \text{m}\cdot\text{s}^{-1} $ here.
$ a = \frac{1}{2}, b = -1 $: This leaves $ \text{kg}^{-1/2} $ unaccounted for, so the units again fail to reduce to plain metres per second.
$ a = 1, b = \frac{1}{2} $: This leaves $ \text{kg}^{3/2} $ uncancelled, the largest mismatch of the four, so it is not consistent either.

Working purely in kilograms, metres and seconds, the only way every kilogram cancels and the metres and seconds reduce to $ \text{m}\cdot\text{s}^{-1} $ is with $ a = \dfrac{1}{2} $ and $ b = -\dfrac{1}{2} $, the square-root-of-tension over square-root-of-mass-per-length combination described by the first option.

So the correct answer is $ a = \frac{1}{2}, b = -\frac{1}{2} $, the first option.

Show Hint

The Correct Option is A

Approach Solution - 1

Approach Solution -2

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