In series combination:
$\frac{1}{k_{s}} = \frac{1}{2k_{1}} + \frac{1}{k_{2}}$
$k_{s} = \left[\frac{1}{2k_{1}} + \frac{1}{k_{2}}\right]^{-1}$
$v = A\omega = 5\pi\text{ cm/s}$
$a = -16\pi^{2}x$
Standard equation of SHM is $A = -\omega^{2}x$
Hence, comparing two equations, we get:
$\omega = 4\pi$
$T = \frac{2\pi}{\omega} = \frac{2\pi}{4\pi} = \frac{1}{2}\text{ sec}.$
Detailed solution not required.
Potential energy $U = mV$
$U = (50x^{2} + 100)10^{-2}$
$F = -\frac{dU}{dx} = -(100x)10^{-2}$
$m\omega^{2}x = -(100 \times 10^{-2})x$
$10 \times 10^{-3}\omega^{2}x = 100 \times 10^{-2}x$
$\omega^{2} = 100 \implies \omega = 10$
$f = \frac{\omega}{2\pi} = \frac{10}{2\pi} = \frac{5}{\pi}$
$F_{max} = m\omega^{2}A = 50\text{ N}$
Given $\frac{1}{2}kx^{2} = \frac{1}{2}\left[\frac{1}{2}kA^{2}\right] \implies x = \frac{A}{\sqrt{2}}$
$F = m\omega^{2}x = \frac{m\omega^{2}A}{\sqrt{2}} = \frac{50}{\sqrt{2}} = 25\sqrt{2}\text{ N}$
$2\pi f = \omega$
$2\pi\sqrt{\frac{ka}{\pi m}} = \omega$
Potential energy $U = \frac{1}{2}m\omega^{2}x^{2} = 2\pi kax^{2}$
$V = 300\text{ m/s}$
$f = 600\text{ Hz}$
$V = f\lambda \implies \lambda = 0.5\text{ m}$
At the distance between the point of maximum amplitude (displacement antinode and displacement node) $= \frac{\lambda}{4} = \frac{0.5}{4} = \frac{1}{8}\text{ m}$
Here $f = \frac{1}{2\pi}\sqrt{\frac{225\pi^{2}}{0.01}} = \frac{1}{2\pi} \times 150\pi = 75\text{ Hz}$.
Thus no resonance.
$\phi_{1} = \frac{5T}{4} \times \frac{2\pi}{T} = \frac{5\pi}{2}$
$\phi_{2} = \frac{5T}{4} \times \frac{2\pi}{5T} = \frac{\pi}{2}$ (or $2\pi$ effective shift)
$\Delta\phi = \frac{5\pi}{2} - 2\pi = \frac{\pi}{2}$
Detailed solution not required.
Here velocity, $v = \omega\sqrt{A^{2} - x^{2}}$
or $v \propto \frac{1}{T}$
$\frac{v_{1}}{v_{2}} = \frac{T_{2}}{T_{1}} = \frac{6}{3} = \frac{2}{1}$
Total energy $E = \frac{1}{2}m\omega^{2}A^{2} \implies E \propto A^{2}$
$T = 2\pi\sqrt{\frac{l}{g}}$
and $T^{\prime} = 2\pi\sqrt{\frac{l}{g + a}}$
$\implies T^{\prime} = 2\pi\sqrt{\frac{l}{g + \frac{g}{3}}}$
$T^{\prime} = 2\pi\sqrt{\frac{3l}{4g}} = \sqrt{\frac{3}{4}}T$
Here $E_{1} = \frac{1}{2}kx^{2}$ or $\sqrt{E_{1}} = \sqrt{\frac{k}{2}}x$
Also $\sqrt{E_{2}} = \sqrt{\frac{k}{2}}y$
Now $E = \frac{1}{2}k(x+y)^{2}$
or $\sqrt{E} = \sqrt{\frac{k}{2}}(x+y) = \sqrt{\frac{k}{2}}x + \sqrt{\frac{k}{2}}y$
$\sqrt{E} = \sqrt{E_{1}} + \sqrt{E_{2}}$
$t_{1} = 2\pi\sqrt{\frac{m}{k_{1}}}$ & $t_{2} = 2\pi\sqrt{\frac{m}{k_{2}}}$
$T = 2\pi\sqrt{m\left(\frac{1}{k_{1}} + \frac{1}{k_{2}}\right)}$
$\therefore T^{2} = 4\pi^{2}\frac{m}{k_{1}} + 4\pi^{2}\frac{m}{k_{2}}$
$T^{2} = t_{1}^{2} + t_{2}^{2}$
We know that, $v = \sqrt{\frac{T}{\mu}}$
$\mu = \frac{m}{l} = \frac{2.5}{20} = 0.125\text{ kg/m}$
Hence wave speed $v = \sqrt{\frac{200}{0.125}} = \sqrt{1600}\text{ m/s} = 40\text{ m/s}$
$T = \frac{20}{40} = 0.5\text{ Second}$
Detailed solution not required.
Detailed solution not required.
Frequency $(\nu) = 800\text{ Hz}$
As the pipe is closed at one end, so
$l_{3} - l_{2} = l_{2} - l_{1} = \frac{\lambda}{2} = 21.5\text{ cm}$
$\lambda = 43.0\text{ cm}$
As $v = \nu\lambda \implies v = \frac{800 \times 43}{100} = 344\text{ ms}^{-1}$
$e = \frac{l_{2} - 3l_{1}}{2}$
First harmonic is obtained at
$l = \frac{\lambda}{4} = 50\text{ cm}$
Third harmonic is obtained for resonance,
$l^{\prime} = \frac{3\lambda}{4} = 3 \times 50 = 150\text{ cm}$
Error in watch = Time taken by sound to reach the man
$= \frac{2000}{330} = \frac{200}{33} \approx 6.06\text{ s}$
Time period, $T = \frac{2\pi}{\omega}$
from given eqn. $\omega = 3.0\text{ s}^{-1}$ or $T = \frac{2\pi}{3} = 2.09\text{ s}$
$I = \frac{P}{4\pi r^{2}} = \frac{4}{4\pi(200)^{2}} \approx 8 \times 10^{-6}\text{ W/m}^{2}$
Detailed solution not required.
$\Delta\phi = \frac{\pi}{2}$
$\frac{I_{1}}{I_{2}} = \left(\frac{A}{2A}\right)^{2} = \frac{1}{4}$
$I_{max} = (\sqrt{I_{1}} + \sqrt{I_{2}})^{2}$
$I_{min} = (\sqrt{I_{1}} - \sqrt{I_{2}})^{2}$
Now put the values:
$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{4}{5}$
$A\omega < \frac{\omega}{k}$
$A < \frac{\lambda}{2\pi}$
Detailed solution not required.
$\frac{2\pi}{\lambda} = \frac{\pi}{3} \implies \lambda = 6$
Distance between nodes $= \frac{\lambda}{2} = \frac{6}{2} = 3\text{ cm}.$
For first resonance
$l_{1} = \frac{330}{500 \times 4}\text{ m} = \frac{330 \times 100}{2000}\text{ cm} = 16.5\text{ cm}$
$100 = 16.5 + (n-1)33$
$n-1 = \frac{100 - 16.5}{33} = \frac{100}{33} - \frac{1}{2}$
$n = \frac{100}{33} + \frac{1}{2} = 3.3 + 0.5 = 3.8$
$y = y_{1} + y_{2}$
$= A[\sin(kx - \omega t)] + A\sin(kx + \omega t)$
$= 2A\sin(kx)\cos(\omega t)$
Wave length $\lambda = \frac{2\pi}{k}$
Distance between adjacent nodes $= \frac{\lambda}{2} = \frac{\pi}{k}$
$R = A\frac{\sin\left(\frac{3 \times 60^{\circ}}{2}\right)}{\sin\left(\frac{60^{\circ}}{2}\right)} = A\frac{\sin 90^{\circ}}{\sin 30^{\circ}} = 2A$
Detailed solution not required.
$\frac{T}{4} = 0.170\text{ s}$
$T = 0.170 \times 4$
$f = \frac{1}{T} = \frac{1}{4 \times 0.17} = \frac{100}{68} = 1.47\text{ Hz}$
Detailed solution not required.
Detailed solution not required.
Detailed solution not required.
$8\sqrt{3} = 4\sqrt{4^{2} - x^{2}}$
$12 = 16 - x^{2}$
$x^{2} = 4 \implies x = 2\text{ cm}$
Detailed solution not required.
Detailed solution not required.
Detailed solution not required.
$3 = 2\pi\sqrt{\frac{500 + 400}{k}}$
$T = 2\pi\sqrt{\frac{400}{k}}$
$\frac{T}{3} = \frac{\sqrt{400}}{\sqrt{900}} = \frac{20}{30} \implies T = 2\text{ s}$
$x(t) = B\cos\left(\frac{\pi}{2} - \frac{\pi}{15}t\right)$
$= B\sin\left(\frac{\pi t}{15}\right)$
$n_{1} = \frac{1}{2 \times 0.6}\sqrt{\frac{400}{0.01}} = \frac{500}{3}$
$n_{1}L_{1} = n_{2}L_{2} = n_{3}L_{3}$
$n_{2} = \frac{1000}{3}$
$n_{3} = \frac{1500}{3}$
Common minimum frequency $= 1000\text{ Hz}$