6 questions · self-marked practice — reveal the answer and mark yourself.
$\text{y}=5\cos(4\times)\sin(20\text{t})$
$\text{y}=10\cos[(252-250)\pi\text{t}]\cos[(252+250)\pi\text{t}]$
Let
$\text{n}_1>\text{n}_2$Beat frequency
$\text{f}_\text{beat}=\text{n}_1-\text{n}_2$$\therefore$ Time period of beats $\text{T}_\text{beats}=\frac{1}{\text{f}_\text{beat}}=\frac{1}{\text{n}_1-\text{n}_2}$
This time period will be wqual to the time interval between successive mixima.
$\text{y}=2\cos(20\pi\text{t}-0.0016\pi\text{x}+7.0\pi)$
Wave is propagated in $+\text{x}$ direction because $\omega\text{t}$ and kx are in with opposite sign standard equation $\text{y}=\text{a}\cos(\omega\text{t}-\text{kx}+\phi)$
a = 2, $\omega=20\pi,\ \text{k}=0.016\pi$ and $\phi=7\pi$
Path difference p =4 m (given) = 400cm
Phase difference $\Delta\phi=\frac{2\pi}{\lambda}\times\text{p}=\frac{2\pi}{\lambda}\times400$
$\Delta\phi=\text{k}\times400=0.016\pi\times400$
Phase difference $\Delta\phi=6.4\pi$ rad.
$\therefore\frac{\text{v}_\text{T}}{\text{v}_0}=\sqrt{\frac{\text{T}_\text{T}}{\text{T}_0}}=\sqrt{\frac{\text{T}_\text{T}}{273}}$
But it is given, $\frac{\text{v}_\text{T}}{\text{v}_0}=\frac{3}{1}$$\Rightarrow\frac{3}{1}=\sqrt{\frac{\text{T}_\text{T}}{\text{T}_0}}\Rightarrow\frac{\text{T}_\text{T}}{273}=9$
$\therefore\text{T}_\text{T}=273\times9=2457\text{k}$
$=2457-273=2184^\circ\text{C}$
$\text{y}=5\sin(100\pi\text{t}-0.4\pi\text{t}+0)$
Wavelength $\lambda,\text{k}=\frac{2\pi}{\lambda}$$\text{k}=0.4\pi$
$\lambda=\frac{2\pi}{\text{k}}=\frac{2\times\pi}{0.4\pi}=5\text{m}$