Question

A string fixed at both the ends, forms a standing wave with node separation of \(5\ \text{cm}\). If the velocity of the wave on the string is \(2\ \text{m/s}\), then the frequency of vibration of the string is

• A. \(40\ \text{Hz}\)
• B. \(20\ \text{Hz}\)
• C. \(10\ \text{Hz}\)
• D. \(0.2\ \text{Hz}\)

Hint:

Adjacent nodes are half a wavelength apart, then use \(v=f\lambda\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In a standing wave the distance between adjacent nodes is \(\dfrac{\lambda}{2}\).

Step 2: Wavelength
\(\dfrac{\lambda}{2}=5\ \text{cm}\), so \(\lambda=10\ \text{cm}=0.1\ \text{m}\).

Step 3: Calculation
\(f=\dfrac{v}{\lambda}=\dfrac{2}{0.1}=20\ \text{Hz}\)

Final Answer:
The string vibrates at 20 Hz. \[ \boxed{\text{(B) }20\ \text{Hz}} \]