Question

The figure shows a snap shot of a vibrating string at t = 0. The particle P is observed moving up with velocity 20 cm/s. The tangent at P makes an angle ${60}^{\circ }$ with x-axis.

The equation of the wave is (in SI units)

• A. $y=(4\times {10}^{-2})sin(10\pi t-50x+\frac{\pi }{4})$
• B. $y=(4\times {10}^{-2})sin(10\pi t+50x-\frac{\pi }{4})$
• C. $y=(4\times {10}^{-2})sin(50x-10\pi t-\frac{\pi }{4})$
• D. $y=(4\times {10}^{-2})sin(10\pi t+50x+\frac{\pi }{4})$

Answer 1

The correct option is D $y=(4\times {10}^{-2})sin(10\pi t+50x+\frac{\pi }{4})$

From graph, $A=4\times {10}^{-2}m$, $\lambda =(5.5-1.5)=4\times {10}^{-2}m$
$\therefore K=\frac{2\pi }{\lambda }=50\pi ,\omega =Kv=50\pi \times \left(\frac{1}{5}\right)=10\pi$
Hence equation of wave is $y=(4\times {10}^{-2})sin(10\pi t+50\pi x+\varphi )$
At x = 0, $y=2\sqrt{2}\times {10}^{-2}\therefore ,2\sqrt{2}\times {10}^{-2}=4\times {10}^{-2}sin\varphi$
$\Rightarrow sin\varphi =\frac{1}{\sqrt{2}},\varphi =\frac{\pi }{4},\frac{3\pi }{4}$
As the particle is moving up at t = 0, x = 0, hence $\varphi =\frac{\pi }{4}\therefore \left(d\right)$