1. Form of a stationary wave on the string

A stationary wave can be written as $y(x,t)=A\sin(kx)\cos(\omega t)$ where $\omega = vk$ for a string wave speed $v=100\,\text{m s}^{-1}$.

Also, the end at $x=0$ is fixed, so $y(0,t)=0$ which is already satisfied by the factor $\sin(kx)$.

The other end at $x=L=3\,\text{m}$ is not fixed; it is being driven in the transverse direction. Therefore, for a stationary wave to exist, $x=3$ need not be a node. The driven end can be either:

• an antinode, or
• any point with nonzero oscillation.

The standard way to get a stationary wave with one end fixed and the other vibrating is that the fixed end must be a node, and the driven end must not be a node. So we need: $\sin(3k) \neq 0$

In addition, the given $\omega$ and $k$ in each option must satisfy $\omega = vk = 100k.$

2. Check each option

Option A

$y=A\sin\left(\frac{\pi x}{6}\right)\cos\left(\frac{50\pi t}{3}\right)$

Here, $k=\frac{\pi}{6}$ so required angular frequency is $\omega=vk=100\cdot \frac{\pi}{6}=\frac{50\pi}{3}$ which matches the given time factor.

Now check at $x=3$: $\sin(3k)=\sin\left(3\cdot \frac{\pi}{6}\right)=\sin\left(\frac{\pi}{2}\right)=1\neq 0$ So the driven end vibrates with maximum amplitude; this is allowed.

✅ A is possible.

Option B

$y=A\sin\left(\frac{\pi x}{3}\right)\cos\left(\frac{100\pi t}{3}\right)$

Here, $k=\frac{\pi}{3}$ so required angular frequency is $\omega=100\cdot \frac{\pi}{3}=\frac{100\pi}{3}$ which matches.

Now check at $x=3$: $\sin(3k)=\sin\left(3\cdot \frac{\pi}{3}\right)=\sin(\pi)=0$ Thus $x=3$ is a node.

But the problem says the other end is vibrating in the $y$-direction, so it cannot be permanently at rest.

❌ B is not possible.

Option C

$y=A\sin\left(\frac{5\pi x}{6}\right)\cos\left(\frac{250\pi t}{3}\right)$

Here, $k=\frac{5\pi}{6}$ so required angular frequency is $\omega=100\cdot \frac{5\pi}{6}=\frac{250\pi}{3}$ which matches.

At $x=3$: $\sin(3k)=\sin\left(3\cdot \frac{5\pi}{6}\right)=\sin\left(\frac{5\pi}{2}\right)=1\neq 0$ So the driven end vibrates; allowed.

✅ C is possible.

Option D

$y=A\sin\left(\frac{5\pi x}{2}\right)\cos(250\pi t)$

Here, $k=\frac{5\pi}{2}$ so required angular frequency is $\omega=100\cdot \frac{5\pi}{2}=250\pi$ which matches.

At $x=3$: $\sin(3k)=\sin\left(3\cdot \frac{5\pi}{2}\right)=\sin\left(\frac{15\pi}{2}\right)=-1\neq 0$ So the end at $x=3$ vibrates; allowed.

✅ D is possible.

3. Final selection

The possible stationary waveforms are: $\boxed{A,\ C,\ D}$

4. Comparison with stored answer

Stored correct answer: A, C, D

Our derived answer matches exactly.