b) Guitar string of length 1.5\ m is made of steel of density 7800\ kg/m^3 and Young's modulus 2 10^11\ N/m^2 . It produces an elastic strain of 1\% in the string. Calculate [3] i) stress developed…

b) Guitar string of length $1.5\mathrm{m}$ is made of steel of density $7800kg/{\mathrm{m}}^3$ and Young's modulus $2\times 10^{11}N/{\mathrm{m}}^2$. It produces an elastic strain of $1\%$ in the string. Calculate

i) Stress developed in the string

Solution: Stress ($\sigma$) is defined as the ratio of the force applied ($F$) to the cross-sectional area ($A$) of the material. It is also related to strain ($ϵ$) and Young's modulus ($Y$) by the formula: $$\sigma =Y\cdot ϵ$$

Given:

• Young's modulus, $Y=2\times 10^{11}N/{\mathrm{m}}^2$
• Elastic strain, $ϵ=1\%=0.01$

Substitute the values: $$\sigma =(2\times 10^{11}N/{\mathrm{m}}^2)\times 0.01$$ $$\sigma =2\times 10^{9}N/{\mathrm{m}}^2$$

Answer: The stress developed in the string is $2\times 10^{9}N/{\mathrm{m}}^2$.

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ii) Frequency of second mode of vibration

Solution: The frequency of vibration of a stretched string is given by: $$f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$$ where:

• $n$ is the mode number,
• $L$ is the length of the string,
• $T$ is the tension in the string,
• $\mu$ is the linear mass density of the string.

First, we need to find the tension $T$ in the string. We know that: $$\sigma =\frac{T}{A}$$ where $A$ is the cross-sectional area of the string. However, we do not have the area $A$, but we can express the tension in terms of stress and area: $$T=\sigma \cdot A$$

The linear mass density $\mu$ is given by: $$\mu =\frac{\text{mass}}{\text{length}}=\frac{\rho \cdot V}{L}=\frac{\rho \cdot A\cdot L}{L}=\rho \cdot A$$ where $\rho$ is the density of the material.

Substitute $\mu$ and $T$ into the frequency equation: $$f_n=\frac{n}{2L}\sqrt{\frac{\sigma \cdot A}{\rho \cdot A}}=\frac{n}{2L}\sqrt{\frac{\sigma }{\rho }}$$

For the second mode of vibration ($n=2$): $$f_2=\frac{2}{2\times 1.5\mathrm{m}}\sqrt{\frac{2\times 10^{9}N/{\mathrm{m}}^2}{7800kg/{\mathrm{m}}^3}}$$ $$f_2=\frac{1}{1.5}\sqrt{\frac{2\times 10^9}{7800}}$$ $$f_2=\frac{1}{1.5}\sqrt{2.5641\times 10^5}$$ $$f_2=\frac{1}{1.5}\times 506.37\mathrm{Hz}$$ $$f_2=337.58\mathrm{Hz}$$

Answer: The frequency of the second mode of vibration is $337.58\mathrm{Hz}$.

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c) The given figure shows standing wave formed in an open organ pipe.

i) Which mode of vibration is shown?

Solution: An open organ pipe supports standing waves with antinodes at both ends. The fundamental frequency (first harmonic) has a single loop, the second harmonic has two loops, and so on. The figure shows three antinodes, which corresponds to the third harmonic (or second overtone).

Answer: The mode of vibration shown is the third harmonic.

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ii) Obtain the frequency of vibration for this mode.

Solution: The frequency of the $n$th harmonic in an open organ pipe is given by: $$f_n=\frac{nv}{2L}$$ where:

• $n$ is the harmonic number,
• $v$ is the speed of sound in air (assumed to be $343m/s$ at room temperature),
• $L$ is the length of the pipe.

For the third harmonic ($n=3$): $$f_3=\frac{3\times 343m/s}{2L}$$

Assuming the length of the pipe $L$ is not given, we cannot calculate the exact frequency. However, if we assume a typical length for an organ pipe (e.g., $L=1.0\mathrm{m}$), then: $$f_3=\frac{3\times 343}{2\times 1.0}=514.5\mathrm{Hz}$$

Note: Since the length $L$ is not provided in the question, the answer should be expressed in terms of $L$: $$f_3=\frac{3\times 343}{2L}$$

Answer: The frequency of vibration for this mode is $\frac{1029}{2L}\mathrm{Hz}$, where $L$ is the length of the pipe.

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Figure for Standing Wave in Open Organ Pipe

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OR

a) Prove that bright and dark fringes are equally spaced in Young's double slit experiment.

Solution: In Young's double slit experiment, light passing through two slits $S_1$ and $S_2$ interferes to produce a pattern of bright and dark fringes on a screen. The path difference between the two waves arriving at a point on the screen determines whether constructive or destructive interference occurs.

Let:

• $d$ be the separation between the two slits,
• $D$ be the distance from the slits to the screen,
• $\lambda$ be the wavelength of light,
• $y$ be the distance from the central bright fringe to a point on the screen.

The path difference ($\Delta x$) between the waves from the two slits is given by: $$\Delta x=\frac{yd}{D}$$

For bright fringes (constructive interference), the path difference is an integer multiple of the wavelength: $$\Delta x=m\lambda (m=0,\pm 1,\pm 2,\dots )$$ $$\frac{yd}{D}=m\lambda$$ $$y=\frac{m\lambda D}{d}$$

For dark fringes (destructive interference), the path difference is an odd multiple of half the wavelength: $$\Delta x=\left(m+\frac{1}{2}\right)\lambda (m=0,\pm 1,\pm 2,\dots )$$ $$\frac{yd}{D}=\left(m+\frac{1}{2}\right)\lambda$$ $$y=\frac{\left(m+\frac{1}{2}\right)\lambda D}{d}$$

The spacing between adjacent bright fringes (or dark fringes) is the difference in $y$ for consecutive values of $m$:

• For bright fringes: $$\Delta y_{\text{bright}}=\frac{(m+1)\lambda D}{d}-\frac{m\lambda D}{d}=\frac{\lambda D}{d}$$
• For dark fringes: $$\Delta y_{\text{dark}}=\frac{\left(m+\frac{3}{2}\right)\lambda D}{d}-\frac{\left(m+\frac{1}{2}\right)\lambda D}{d}=\frac{\lambda D}{d}$$

Thus, the spacing between adjacent bright fringes and adjacent dark fringes is the same: $$\Delta y_{\text{bright}}=\Delta y_{\text{dark}}=\frac{\lambda D}{d}$$

Conclusion: Bright and dark fringes are equally spaced in Young's double slit experiment.

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b) A parallel beam of monochromatic light is incident on diffraction grating having $500lines/mm$ to get second order maxima at $30^{\circ }$. Calculate the wavelength of light used.

Solution: A diffraction grating has $N$ lines per unit length. The grating equation for maxima is: $$d\sin \theta =m\lambda$$ where:

• $d$ is the spacing between adjacent slits,
• $\theta$ is the angle of diffraction,
• $m$ is the order of the maxima,
• $\lambda$ is the wavelength of light.

Given:

• Number of lines per mm = $500lines/mm$
• Spacing between lines, $d=\frac{1}{500}\mathrm{mm}=2\times 10^{-6}\mathrm{m}$
• Angle of second order maxima, $\theta =30^{\circ }$
• Order of maxima, $m=2$

Substitute the values into the grating equation: $$2\times 10^{-6}\sin 30^{\circ }=2\lambda$$ $$2\times 10^{-6}\times 0.5=2\lambda$$ $$1\times 10^{-6}=2\lambda$$ $$\lambda =\frac{1\times 10^{-6}}{2}=5\times 10^{-7}\mathrm{m}$$

Answer: The wavelength of light used is $500\mathrm{nm}$.

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c) State and prove Brewster's law.

Solution: Statement of Brewster's Law: When unpolarized light is incident on a transparent dielectric surface at a particular angle of incidence (called Brewster's angle or polarization angle), the reflected light is completely plane-polarized. Brewster's law states that: $$\tan {\theta }_p=\frac{n_2}{n_1}$$ where:

• ${\theta }_p$ is Brewster's angle,
• $n_1$ is the refractive index of the first medium (incident medium),
• $n_2$ is the refractive index of the second medium (transmitted medium).

Proof: Consider unpolarized light incident on a boundary between two media with refractive indices $n_1$ and $n_2$. Let ${\theta }_i$ be the angle of incidence and ${\theta }_r$ be the angle of refraction.

According to Snell's law: $$n_1\sin {\theta }_i=n_2\sin {\theta }_r$$

For the reflected light to be completely plane-polarized, the angle between the reflected ray and the refracted ray must be $90^{\circ }$. This means: $${\theta }_i+{\theta }_r=90^{\circ }$$ $${\theta }_r=90^{\circ }-{\theta }_i$$

Substitute ${\theta }_r$ into Snell's law: $$n_1\sin {\theta }_i=n_2\sin (90^{\circ }-{\theta }_i)$$ $$n_1\sin {\theta }_i=n_2\cos {\theta }_i$$ $$\frac{\sin {\theta }_i}{\cos {\theta }_i}=\frac{n_2}{n_1}$$ $$\tan {\theta }_i=\frac{n_2}{n_1}$$

Thus, Brewster's angle ${\theta }_p$ satisfies: $$\tan {\theta }_p=\frac{n_2}{n_1}$$

Conclusion: Brewster's law is proven.