The equation of a wave travelling on a string is y = sin[20πx + 10πt], where x and t are distance and time in SI units. The minimum distance between two points having the same oscillating speed is:

❓ Question

The equation of a wave travelling on a string is

$$y=\sin (20\pi x+10\pi t),$$

where $x$ and $t$ are distance and time in SI units.
The minimum distance between two distinct points on the string having the same oscillating speed (transverse velocity) is: (find it).

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🖼️ Question Image

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✍️ Short Solution

We interpret “oscillating speed” as the instantaneous transverse velocity of a point of the string, i.e. $\dfrac{\partial y}{\partial t}$.

Given

$$y(x,t)=\sin (20\pi x+10\pi t)\mathrm{.}$$

Compute transverse velocity:

$$v(x,t)=\frac{\mathrm{\partial }y}{\mathrm{\partial }t}=10\pi \cos (20\pi x+10\pi t)\mathrm{.}$$

Two points $x_1$ and $x_2$ have the same transverse velocity at the same instant $t$ when

$$\cos (20\pi x_1+10\pi t)=\cos (20\pi x_2+10\pi t)\mathrm{.}$$

Using the cosine identity $\cos A=\cos B$, we have two families of solutions:

(i) $20\pi x_1 + 10\pi t = 20\pi x_2 + 10\pi t + 2\pi n$
$\Rightarrow 20\pi (x_1-x_2)=2\pi n$
$\Rightarrow x_1-x_2=\dfrac{n}{10}$

(ii) $20\pi x_1 + 10\pi t = -\big(20\pi x_2 + 10\pi t\big) + 2\pi n$
$\Rightarrow 20\pi(x_1+x_2) + 20\pi t = 2\pi n$
$\Rightarrow x_1+x_2 = \dfrac{n - 10t}{10}.$

The second family depends on time $t$ and allows one to pick points arbitrarily close (by appropriate choice of $x_1$), so it does not give a fixed minimal spatial separation independent of $t$. The physically meaningful, time-independent minimal separation between distinct points that always share the same instantaneous transverse velocity arises from family (i).

From (i), the spatial separations are

$$\Delta x = |x_1 - x_2| = \frac{n}{10},\qquad n\in\mathbb{Z}.$$

The smallest non-zero separation occurs at $n=1$:

$$\Delta x_{\min} = \frac{1}{10}\ \text{m} = 0.1\ \text{m}.$$

Note: the wavelength $\lambda$ of this wave is

$$\lambda =\frac{2\pi }{k}=\frac{2\pi }{20\pi }=\frac{1}{10}\text{ m},$$

so the minimum separation equals one wavelength $\lambda$.

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🖼️ Image Solution

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✅ Conclusion & Video Solution

• Transverse velocity: $v(x,t)=10\pi\cos(20\pi x+10\pi t)$
  

• Condition $\cos(\cdot)=\cos(\cdot)$ yields $\Delta x = n/10$
  

• Smallest nonzero distance between distinct points with the same instantaneous oscillating speed is

$$\boxed{{\displaystyle 0.1\text{ m}}}\mathrm{.}$$