Question

A standing wave exists in a string of length 150 cm which is fixed at both ends with rigid supports. The displacement amplitude of a point at a distance of 10 cm from one of the ends is $5\sqrt{3}$ mm. The distance between the two nearest points, within the same loop and having displacement amplitude equal to $5\sqrt{3}$ mm is 10 cm. The maximum displacement amplitude of the particles in the string is

• A. None of these
• B. 10 mm
• C. 15 mm
• D. 20 mm

Answer 1

The correct option is B

10 mm

$y=Asinkxcos\omega t$

$5\sqrt{3}=Asink\times 10$

$5\sqrt{3}=Asink\times 20$

After solving $k=\frac{\pi }{30}$

$5\sqrt{3}=Asin(\frac{\pi }{30}\times 10)\Rightarrow A=10mm$