The equations of two simple harmonic waves ar | vedclass

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(D) The standard equation for a wave is $Y = A \sin 2 \pi \left(\frac{t}{T} - \frac{x}{\lambda}\right)$.For the first wave: $Y_1 = 2 \sin 2 \pi \left(

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same velocity

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(D) The standard equation for a wave is $Y = A \sin 2 \pi \left(\frac{t}{T} - \frac{x}{\lambda}ight)$.For the first wave: $Y_1 = 2 \sin 2 \pi \left(\frac{4t}{0.2} - \frac{4x}{2}ight) = 2 \sin 2 \pi \left(\frac{t}{0.05} - \frac{x}{0.5}ight)$.Here,$T_1 = 0.05 	ext{ s}$ and $\lambda_1 = 0.5 	ext{ m}$.Velocity $v_1 = \frac{\lambda_1}{T_1} = \frac{0.5}{0.05} = 10 	ext{ m/s}$.For the second wave: $Y_2 = 4 \sin 2 \pi \left(\frac{4t}{0.16} - \frac{4x}{1.6}ight) = 4 \sin 2 \pi \left(\frac{t}{0.04} - \frac{x}{0.4}ight)$.Here,$T_2 = 0.04 	ext{ s}$ and $\lambda_2 = 0.4 	ext{ m}$.Velocity $v_2 = \frac{\lambda_2}{T_2} = \frac{0.4}{0.04} = 10 	ext{ m/s}$.Since $v_1 = v_2 = 10 	ext{ m/s}$,both waves have the same velocity.

The equations of two simple harmonic waves are given by $Y_1 = 2 \sin 8 \pi \left(\frac{t}{0.2} - \frac{x}{2}\right) \text{ m}$ and $Y_2 = 4 \sin 8 \pi \left(\frac{t}{0.16} - \frac{x}{1.6}\right) \text{ m}$. Then both waves have:

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