Q.The transverse displacement of a string (clamped at its both ends) is given by

[!FORMULA] $y(x, t) = 0.06 \sin\left(\frac{2\pi}{3}\,x\right) \cos(120\,\pi t)$

where $x$ and $y$ are in m and $t$ in s. The length of the string is $1.5\ \text{m}$ and its mass is $3.0 \times 10^{-2}\ \text{kg}$.
Answer the following:

Concept: Wave Speed on String

The given function $y(x,t) = 0.06 \sin\left(\frac{2\pi}{3}x\right) \cos(120\pi t)$ is a product of a space-dependent sine and a time-dependent cosine — this is the standard form of a stationary (standing) wave.

(a) It represents a stationary wave, because the $x$ and $t$ parts are separated into a product, not a function of $(x \pm vt)$.

(b) Using the identity $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$, we write:

This is the sum of two travelling waves: $y_1 = 0.03 \sin\left(\frac{2\pi}{3}x - 120\pi t\right)$ (rightward) and $y_2 = 0.03 \sin\left(\frac{2\pi}{3}x + 120\pi t\right)$ (leftward).

Comparing with $y = A \sin(kx \mp \omega t)$:

• Wave number $k = \frac{2\pi}{3} \ \text{rad/m}$, so wavelength $\lambda = \frac{2\pi}{k} = 3\ \text{m}$.
• Angular frequency $\omega = 120\pi \ \text{rad/s}$, so frequency $f = \frac{\omega}{2\pi} = 60\ \text{Hz}$. …

The equation $y(x,t) = 0.06 \sin\left(\frac{2\pi}{3}x\right) \cos(120\pi t)$ is of the form $y = 2A \sin(kx) \cos(\omega t)$. This is the classic signature of a stationary (standing) wave — the $x$ and $t$ parts are separated into a product, not a function of $(x \pm vt)$. The string is clamped at both ends, so the ends are nodes, and the wave pattern oscillates in place without travelling.

1. Identify the wave type

   A travelling wave has the form $f(kx \pm \omega t)$. Here, the variables are separated: $\sin(kx)$ times $\cos(\omega t)$. That is a standing wave.

   Watch out

   A common mistake is to see the product and think it's a travelling wave. The key test: can you rewrite it as $f(kx - \omega t)$ or $f(kx + \omega t)$? If not, it's stationary.

2. Extract wave parameters

   Compare with the standard standing wave form $y = 2A \sin(kx) \cos(\omega t)$.

   • $k = \frac{2\pi}{3}$ (from $\sin(\frac{2\pi}{3}x)$)
   • $\omega = 120\pi$ (from $\cos(120\pi t)$)
   • Amplitude of each travelling component: $2A = 0.06 \Rightarrow A = 0.03\ \text{m}$

3. Find wavelength, frequency, and speed of each travelling wave

   For a standing wave formed by superposition of two identical waves moving oppositely:

   • Wavelength: $k = \frac{2\pi}{\lambda} \Rightarrow \lambda = \frac{2\pi}{k} = \frac{2\pi}{2\pi/3} = 3\ \text{m}$
   • Frequency: $\omega = 2\pi f \Rightarrow f = \frac{\omega}{2\pi} = \frac{120\pi}{2\pi} = 60\ \text{Hz}$
   • Wave speed: $v = f\lambda = 60 \times 3 = 180\ \text{m/s}$ Each travelling wave has the same $\lambda$, $f$, and $v$; one moves in $+x$ direction, the other in $-x$ direction. …

Step 1: Compare $y(x,t)=0.06\sin\left(\frac{2\pi}{3}x\right)\cos(120\pi t)$ against the two diagnostic forms — a travelling wave depends only on $(x\pm vt)$, a stationary wave is a product $f(x)g(t)$. Here space and time are already separated as a product, so this is a stationary wave (answers part a).

Step 2: Split it into its two travelling components using $\sin A\cos B=\frac12[\sin(A+B)+\sin(A-B)]$:

each term has amplitude $0.03\ \text{m}$, one travelling in $+x$, the other in $-x$. …