TG EAMCET 4 May 2025 Shift 1 Paper

The tube is 50 cm long and is closed at one end.Finding the Wavelength:For the fifth harmonic in a closed pipe, the formula is:$\;\frac{5\lambda}{4}=L$This means five-fourths of the wavelength fits into the total length $(L)$ of the tube.Let's solve for wavelength $(\lambda)$ :$\lambda=\;\frac{4L}{5}=\;\frac{4\times50}{5}=40\text{ cm}$Calculating the Phase Difference:The phase difference $(\Delta\varphi)$ between two points separated by a distance $x$ is:$\Delta\varphi=\;\frac{2\pi}{\lambda}\times x$Here, $x=42\text{ cm}$ (the distance from the open end to the other particle) and $\lambda=40\text{ cm}$ :$\Delta\varphi=\;\frac{2\pi}{40}\times42$This can be calculated as:$\Delta\varphi=\;\frac{2\pi\times42}{40}=\;\frac{84\pi}{40}=\;\frac{21\pi}{10}$If you convert this to degrees:$\Delta\varphi=\;\frac{21\pi}{10}\times\;\frac{180^{\circ}}{\pi}=378^{\circ}$ But phases repeat every $360^{\circ}$. So, we subtract $360^{\circ}$ to get the simpler value:$378^{\circ}-360^{\circ}=18^{\circ}$So, the phase difference is $18^{\circ}$.