Question

The equation of a stationary wave along a stretched string is given by \[ y = 5 \sin \left( \frac{\pi x}{3} \right) \cos (40 \pi t) \] Additional Information Here, \(x\) and \(y\) are in cm and \(t\) in seconds. The separation between two adjacent nodes is:

• A. 1.5 cm
• B. 3 cm
• C. 6 cm
• D. 14 cm

Hint:

The separation between two adjacent nodes in a stationary wave is given by the distance between consecutive integer multiples of \( \frac{\pi}{k} \), where \( k \) is the wave number.

Answer 1

The Correct Option is B

We analyze the equation for a stationary wave on a string: \[ y = 5 \sin \left( \frac{\pi x}{3} \right) \cos (40 \pi t) \] The standard form for a stationary wave is: \[ y = A \sin(kx) \cos(\omega t) \] where: - \( A \) is the amplitude, - \( k \) is the wave number, - \( \omega \) is the angular frequency, - \( x \) is position, - \( t \) is time. Comparing the given equation: \[ y = 5 \sin \left( \frac{\pi x}{3} \right) \cos (40 \pi t) \] We find: - \( A = 5 \), - \( k = \frac{\pi}{3} \), - \( \omega = 40 \pi \). Nodes in a stationary wave are points of zero displacement. Nodes occur when the sine term is zero: \[ \sin(kx) = 0 \] This gives: \[ kx = n\pi, \quad n = 0, 1, 2, 3, \dots \] Solving for \( x \): \[ x = \frac{n\pi}{k} = \frac{n\pi}{\frac{\pi}{3}} = 3n \, \text{cm} \] The nodes are at \( x = 0, 3, 6, 9, \dots \). The distance between adjacent nodes (e.g., between \( x = 0 \) and \( x = 3 \)): \[ \text{Separation between nodes} = 3 \, \text{cm} \] Therefore, the answer is 3 cm.