For a transverse wave travelling along a straight line, the distance between two peaks (crests) is 5

Q: Why is the distance between any two crests $n\lambda$?

The distance between two consecutive crests is exactly one wavelength ($\lambda$). Therefore, if you pick any two crests on the wave, the distance between them must be a whole number multiple of $\lambda$, which we represent as $n\lambda$.

Q: Why is the distance between a crest and a trough $(m + 1/2)\lambda$?

A trough is located exactly halfway between two crests. So, the distance from a crest to the very next trough is $\lambda/2$. If you pick any crest and any trough, the distance will be some number of full wavelengths plus that extra half wavelength, giving $(m + 1/2)\lambda$.

Q: Why do we divide the two equations?

We have two equations but three unknowns ($n$, $m$, and $\lambda$). By dividing the equations, we eliminate $\lambda$ and create a single equation relating the two integers $n$ and $m$. This allows us to find valid integer pairs.

Q: How do we know which values of $m$ to pick?

Once we have the equation $3n = 10m + 5$, we know that $n$ must be an integer. This means $10m + 5$ must be divisible by 3. We can simply test integer values for $m$ ($1, 2, 3, \dots$) and see which ones result in a whole number for $n$.

Q: Can $m$ or $n$ be negative?

In this physical context, $n$ and $m$ represent the number of wavelengths between points on the wave. Since distance is positive, $n$ and $m$ must be non-negative integers.

Question

For a transverse wave travelling along a straight line, the distance between two peaks (crests) is $5	ext{ m}$, while the distance between one crest and one trough is $1.5	ext{ m}$. The possible wavelengths (in metre) of the waves are

Options:

(A) $1, 2, 3, \dots$

(B) $1, \frac{1}{3}, \frac{1}{5}, \dots$

(C) $1, 3, 5, \dots$

(D) $\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \dots$

Sigma Percentile · Accepted Answer

### Visualizing the Wave Geometry

Imagine a transverse wave travelling along a straight line. The distance between any two consecutive crests is exactly one wavelength, $\lambda$. Therefore, the distance between *any* two crests must be an integer multiple of $\lambda$. We can write this mathematically as:

$$n\lambda = 5$$

where $n$ is an integer.

### The Crest-to-Trough Constraint

Next, we consider the distance between a crest and a trough. A trough is located exactly halfway between two crests, meaning the distance from a crest to the adjacent trough is $\lambda/2$.

Thus, the distance between *any* crest and *any* trough will be some integer number of full wavelengths plus an extra half wavelength. We can express this as:

$$(m + \frac{1}{2})\lambda = 1.5$$

where $m$ is another integer.

### The Master Equation

To solve for the possible wavelengths, we need to eliminate $\lambda$ and find a relationship between our two integers, $n$ and $m$. Let's first simplify our second equation by multiplying by 2:

$$(2m + 1)\lambda = 3$$

Now, we divide our first equation by this simplified second equation:

$$\frac{n\lambda}{(2m + 1)\lambda} = \frac{5}{3}$$

The $\lambda$ terms cancel out beautifully! Cross-multiplying gives us our master Diophantine equation:

$$3n = 5(2m + 1)$$

$$3n = 10m + 5$$

### Hunting for Integer Solutions

Since $n$ and $m$ represent physical counts of waves, they must be non-negative integers. We can find valid pairs by testing values for $m$ and checking if $n$ results in an integer.

Let's test $m = 1$:

$$3n = 10(1) + 5 = 15 \implies n = 5$$

This is a valid pair! Substituting $n = 5$ back into our very first equation ($n\lambda = 5$), we get:

$$5\lambda = 5 \implies \lambda = 1	ext{ m}$$

Let's find the next valid pair. Testing $m = 2$ and $m = 3$ does not yield an integer for $n$. However, testing $m = 4$ gives:

$$3n = 10(4) + 5 = 45 \implies n = 15$$

Substituting $n = 15$ back into $n\lambda = 5$, we get:

$$15\lambda = 5 \implies \lambda = \frac{1}{3}	ext{ m}$$

Continuing this pattern, the next valid integer is $m = 7$, which gives $n = 25$ and $\lambda = 1/5	ext{ m}$.

**Therefore, the possible wavelengths are $1, \frac{1}{3}, \frac{1}{5}, \dots$ meters.**

Animated Solution for Physics - Waves: For a transverse wave travelling along a straight line, the distance between two peaks (crests) is $5 m$ , while the distance between one crest and one trough is $1.5 m$ . The possible wavelengths (in metre) of the waves are

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Visualized Solution

$nλ = 5$

$(m + \frac{1}{2}) λ = 1.5$

$(2 m + 1) λ = 3$

$\frac{nλ}{( 2 m + 1 ) λ} = \frac{5}{3}$

$3 n = 10 m + 5$

$For m = 1$

$For m = 4, 7, \dots$

The Sigma Insight: Wave Equation and Wave Speed

Visualizing the Wave Geometry

The Crest-to-Trough Constraint

The Master Equation

Hunting for Integer Solutions

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Visualized Solution

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For m=1

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