Minimum Distance for Same Transverse Velocity in Wave

❓ 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).

---

🖼️ Question Image

---

✍️ Short Explanation

This problem is based on:

👉 Wave motion
👉 Particle velocity in wave
👉 Phase difference.

Main idea:

For wave:

$$y=\sin(kx+\omega t)$$

oscillating speed depends on:

$$\left|\frac{\partial y}{\partial t}\right|$$

---

🔷 Step 1 — Find Oscillating Speed 💯

Given wave:

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

Particle velocity:

$$v=\frac{\partial y}{\partial t}$$

Differentiate w.r.t. $t$:

$$v=10\pi\cos(20\pi x+10\pi t)$$

Oscillating speed means magnitude:

$$|v|=10\pi |\cos(20\pi x+10\pi t)|$$

Thus two points have same oscillating speed if:

$$|\cos\phi_1| = |\cos\phi_2|$$

---

🔷 Step 2 — Minimum Phase Difference

Condition:

$$|\cos\phi_1|=|\cos\phi_2|$$

Minimum non-zero phase difference occurs when:

$$\Delta\phi=\pi$$

Now phase difference due to distance:

$$\Delta\phi=k\Delta x$$

where:

$$k=20\pi$$

Thus:

$$20\pi \Delta x=\pi$$
$$\Delta x=\frac1{20}\text{ m}$$
$$=0.05\text{ m}$$
$$=5\text{ cm}$$

---

🔷 Step 3 — JEE Trap Alert 🚨

❌ Oscillating speed aur wave speed confuse kar dena

❌ Magnitude condition ignore kar dena

❌ $k=\frac{2\pi}{\lambda}$ relation bhool jaana

Remember:

$$\boxed{ v_{particle}=\frac{\partial y}{\partial t} }$$

---

✅ Final Answer

$$\boxed{ 5\text{ cm} }$$

(Option 2)

---

📚 Related Topics

• RC Circuit Phase 90° Condition
• MOI of Combined Bodies Using Parallel Axis
• Find Height Using Refraction in Layered Media