Match Standing Wave Systems with Wave Properties and Wavelength

To match each system in Column I with its appropriate wave properties and fundamental wavelength in Column II, we analyze the nature of the waves (longitudinal vs. transverse) and the boundary conditions for each standing wave system:

---

System (A): Pipe closed at one end

1. Wave Nature: Sound waves in air columns are pressure/displacement waves that oscillate parallel to the direction of wave propagation, making them longitudinal waves $\rightarrow$ (p).
2. Fundamental Wavelength ($\lambda_f$): A closed pipe of length $L$ has a displacement node at the closed end ($x = 0$) and a displacement antinode at the open end ($x = L$). For the fundamental mode (lowest natural frequency):
   $L = \frac{\lambda_f}{4} \implies \lambda_f = 4L \rightarrow \mathbf{(t)}$

Thus, (A) $\rightarrow$ (p, t).

---

System (B): Pipe open at both ends

1. Wave Nature: Waves inside the open pipe are sound waves in air, which are longitudinal waves $\rightarrow$ (p).
2. Fundamental Wavelength ($\lambda_f$): An open pipe of length $L$ has displacement antinodes at both open ends ($x = 0$ and $x = L$) and a single node at the center. For the fundamental mode:
   $L = \frac{\lambda_f}{2} \implies \lambda_f = 2L \rightarrow \mathbf{(s)}$

Thus, (B) $\rightarrow$ (p, s).

---

System (C): Stretched wire clamped at both ends

1. Wave Nature: Waves on a stretched string/wire oscillate perpendicular to the length of the string, making them transverse waves $\rightarrow$ (q).
2. Fundamental Wavelength ($\lambda_f$): Clamped fixed ends force displacement nodes at $x = 0$ and $x = L$. The fundamental mode contains a single antinode at the center:
   $L = \frac{\lambda_f}{2} \implies \lambda_f = 2L \rightarrow \mathbf{(s)}$

Thus, (C) $\rightarrow$ (q, s).

---

System (D): Stretched wire clamped at both ends and at mid-point

1. Wave Nature: Standing waves on a stretched wire are transverse waves $\rightarrow$ (q).
2. Fundamental Wavelength ($\lambda_f$): Clamping the wire at $x = 0$, $x = L/2$, and $x = L$ forces fixed displacement nodes at these points, dividing the wire into two independent segments each of length $l = \frac{L}{2}$. For the lowest possible natural frequency of this clamped system, each segment vibrates in its fundamental mode:
   $l = \frac{L}{2} = \frac{\lambda_f}{2} \implies \lambda_f = L \rightarrow \mathbf{(r)}$

Thus, (D) $\rightarrow$ (q, r).

---

Final Matching Summary:

• (A) $\rightarrow$ p, t
• (B) $\rightarrow$ p, s
• (C) $\rightarrow$ q, s
• (D) $\rightarrow$ q, r