- amplitude is $2A$
- velocity is in the negative x-direction
- period is $\frac{T}{\lambda}$
- speed is $\frac{x}{t}$
- speed is $\frac{\lambda}{T}$

Solution :

Assume $A$, $T$, $\lambda$ are positive constants. The equation \[y=A\sin\left[2\pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)\right]\] represents a wave whose:

**Choice 5 is the answer.**

Choice 1 is incorrect because $\sin\theta\leq 1$ for any $\theta$ so $y\leq A$.

To see that choice 2 is incorrect, suppose that we are following the minimum wave as time passes (by changing $x$ with respect to time to a specific $x(t)$) in such a way that the wave seems to stand still (i.e. we are travelling at the same velocity as the wave). For the wave to 'stand still', we need $x(t)$ to satisfy \[2\pi\left(\frac{t}{T}-\frac{x(t)}{\lambda}\right) = const. = C\] Satisfying this relationship ensures that the wave always looks the same to us as: \[y=A\sin\left[2\pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)\right] = A\sin C = constant.\] Since we are following one of the minimums of the wave, we can choose $C = 0$ to get the minimum at $\sin 0$. Thus we get: \[2\pi\left(\frac{t}{T}-\frac{x(t)}{\lambda}\right) = 0 \implies x(t) = \frac{\lambda}{T}t.\] To get the sign of the velocity, we take the derivative of $x(t)$: \[\frac{dx}{dt} = \frac{\lambda}{T}>0\] since $\lambda,T > 0$. This also tells us choice 5 is correct since the speed is $\frac{\lambda}{T}.$

Choice 3 is wrong as the units do not work out.

Choice 4 is wrong because we just calculated the speed to be $\frac{\lambda}{T}.$

- amplitude is $2A$
- velocity is in the negative x-direction
- period is $\frac{T}{\lambda}$
- speed is $\frac{x}{t}$
- speed is $\frac{\lambda}{T}$

Solution :

Choice 1 is incorrect because $\sin\theta\leq 1$ for any $\theta$ so $y\leq A$.

To see that choice 2 is incorrect, suppose that we are following the minimum wave as time passes (by changing $x$ with respect to time to a specific $x(t)$) in such a way that the wave seems to stand still (i.e. we are travelling at the same velocity as the wave). For the wave to 'stand still', we need $x(t)$ to satisfy \[2\pi\left(\frac{t}{T}-\frac{x(t)}{\lambda}\right) = const. = C\] Satisfying this relationship ensures that the wave always looks the same to us as: \[y=A\sin\left[2\pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)\right] = A\sin C = constant.\] Since we are following one of the minimums of the wave, we can choose $C = 0$ to get the minimum at $\sin 0$. Thus we get: \[2\pi\left(\frac{t}{T}-\frac{x(t)}{\lambda}\right) = 0 \implies x(t) = \frac{\lambda}{T}t.\] To get the sign of the velocity, we take the derivative of $x(t)$: \[\frac{dx}{dt} = \frac{\lambda}{T}>0\] since $\lambda,T > 0$. This also tells us choice 5 is correct since the speed is $\frac{\lambda}{T}.$

Choice 3 is wrong as the units do not work out.

Choice 4 is wrong because we just calculated the speed to be $\frac{\lambda}{T}.$

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