The equation of stationary wave along a stretched string is given by y=5 sin (π x)/3 cos 40 π t . Here x and y are in cm and t in second. The separation between two adjacent nodes is

Correct answer: (B) 3 cm. y=5 sin ((π)/3 x ) cos (40 π t), ω=40 π k=3 ~cm, λ=(2 π)/k=6 ~cm Separation between nodes =(λ)/2=3 ~cm

Q147 / 155 · WBJEE 2025 · Physics

The equation of stationary wave along a stretched string is given by $y=5 \sin \frac{\pi x}{3} \cos 40 \pi t$. Here x and y are in cm and t in second. The separation between two adjacent nodes is

A
1.5 cm
Correct answer
B
3 cm
C
6 cm
D
14 cm

Solution

$\mathrm{y}=5 \sin \left(\frac{\pi}{3} \mathrm{x}\right) \cos (40 \pi \mathrm{t}), \omega=40 \pi \mathrm{k}=3 \mathrm{~cm}, \lambda=\frac{2 \pi}{\mathrm{k}}=6 \mathrm{~cm}$ Separation between nodes $=\frac{\lambda}{2}=3 \mathrm{~cm}$

Hence, the correct answer is option: 'B'.

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