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Q15.11(b):
The transverse displacement of a string (clamped at its both ends) is given by $y(x, t) = 0.06 \sin(\frac{2\pi x}{3}) \cos(120 \pi t)$ where $x$ and $y$ are in m and $t$ in s. The length of the string is 1.5 m and its mass is $3.0 \times 10^{-2}$ kg. Answer the following : (b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave ?

Solution :

Step 1: Decomposition of the Standing Wave Equation

The given displacement equation for the standing wave is:

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

Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$, we rewrite the equation to identify the component traveling waves. Let $A = \frac{2\pi x}{3}$ and $B = 120 \pi t$.

Then, $y(x, t) = 0.03 \left[ 2 \sin\left(\frac{2\pi x}{3}\right) \cos(120 \pi t) \right]$

$y(x, t) = 0.03 \left[ \sin\left(\frac{2\pi x}{3} + 120 \pi t\right) + \sin\left(\frac{2\pi x}{3} - 120 \pi t\right) \right]$

[By Trigonometric Identity for the superposition of two waves travelling in opposite directions]

Step 2: Identification of Wave Parameters

The general form of a travelling wave is $y = A_0 \sin(kx \pm \omega t)$. Comparing the terms in the expansion above:

  • The angular wave number $k = \frac{2\pi}{3} \text{ rad/m}$
  • The angular frequency $\omega = 120\pi \text{ rad/s}$

Step 3: Calculating Wavelength ($\lambda$)

The relationship between $k$ and $\lambda$ is $k = \frac{2\pi}{\lambda}$ [By definition of wave number].

$\frac{2\pi}{3} = \frac{2\pi}{\lambda} \implies \lambda = 3 \text{ m}$

Step 4: Calculating Frequency ($\nu$)

The relationship between $\omega$ and $\nu$ is $\omega = 2\pi\nu$ [By circular frequency definition].

$120\pi = 2\pi\nu \implies \nu = \frac{120\pi}{2\pi} = 60 \text{ Hz}$

Step 5: Calculating Wave Speed ($v$)

The speed of each wave is given by $v = \frac{\omega}{k} = \nu \lambda$ [By wave speed-frequency relation].

$v = 60 \text{ Hz} \times 3 \text{ m} = 180 \text{ m/s}$

x=0 x=1.5m Standing Wave Envelope

Step 6: Summary of Results

Parameter Value
Wavelength ($\lambda$) $3 \text{ m}$
Frequency ($\nu$) $60 \text{ Hz}$
Wave Speed ($v$) $180 \text{ m/s}$

Final Solution: The standing wave is a superposition of two waves of wavelength $3 \text{ m}$, frequency $60 \text{ Hz}$, and speed $180 \text{ m/s}$ travelling in opposite directions.


More Questions from Class 11 Physics Waves EXERCISES


CBSE Solutions for Class 11 Physics Waves


Chapters in CBSE - Class 11 Physics


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