Question

A sinusoidal wave represented as $y_i=0.004\cos \pi (4x-100t)\text{m}$ travels along a string towards a boundary at $x=0$ with a second string. If the wave speed on the second string is $50\text{m/s}$ , then which of the following statement(s) is/are correct?

• A. Amplitude of the reflected wave is $\frac{4}{3}\times {10}^{-3}\text{m}$.
• B. Amplitude of the transmitted wave is $\frac{16}{3}\times {10}^{-3}\text{m}$.
• C. Amplitude of the reflected wave is $\frac{16}{3}\times {10}^{-3}\text{m}$.
• D. Amplitude of the transmitted wave is $\frac{4}{3}\times {10}^{-3}\text{m}$.

Answer 1

Answer: (A), (B)

Amplitude of the transmitted wave is $\frac{16}{3}\times {10}^{-3}\text{m}$.

Given that ,
$y_i=0.004\cos \pi (4x-100t)\text{m}......\left(1\right)$
Wave speed on $2^{nd}$ stretched string $\left(v_2\right)=50\text{m/s}$
Wave speed on $1^{st}$ stretched string $\left(v_1\right)=\frac{\omega }{k}$

From $\left(1\right)$, we get $\omega =100\pi \text{rad/s}$, $k=4\pi {\text{m}}^{-1}$ and $A_i=0.004\text{m}$
$\therefore v_1=\frac{100}{4}=25\text{m/s}$.
From this, we can conclude that $v_2>v_1$

Amplitude of the reflected wave is given by
$A_r=|\frac{v_2-v_1}{v_2+v_1}|A_i$

From the given data,
$A_r=\left(\frac{50-25}{50+25}\right)\times 4\times {10}^{-3}=\frac{25}{75}\times 4\times {10}^{-3}$
$\Rightarrow A_r=\frac{4}{3}\times {10}^{-3}\text{m}$

Now, Amplitude of transmitted wave is given by
$A_t=|\frac{2v_2}{v_1+v_2}|A_i$

From the given data,
$A_t=\frac{50\times 2}{50+25}\times 4\times {10}^{-3}=\frac{100}{75}\times 4\times {10}^{-3}$
$\Rightarrow A_t=\frac{16}{3}\times {10}^{-3}\text{m}$

Thus, options (a) and (b) are the correct answers.