Physics M.C.Q (1 Marks)

1. 300

Exaplanation:

Here, $\text{y}=10^{-4}\sin\Big(600\text{t}-2\text{x}+\frac{\pi}{3}\Big)$

Compare it with the standard equation of a travalling wave

$\text{y}=\text{r}\Big(\frac{2\pi\text{t}}{\text{T}}-\frac{2\pi\text{x}}{\lambda}\neq\phi\Big)$

$\frac{2\pi}{\text{T}}=600,\text{T}=\frac{2\pi}{600}=\frac{\pi}{300}\text{s}$

$\frac{2\pi}{\lambda}=+2\lambda =\frac{2\pi}{2}=\pi\text{m}$

$\nu=\frac{\lambda}{\text{T}}=\frac{\pi}{\frac{\pi}{300}}=300\text{m/s}$

Explanation:

When qbserver is at rest and source of sound id moving towards observer then observed frequency n'.

$\text{n}'=\Big(\frac{\text{v}}{\text{V}-\text{v}_\text{s}}\Big)\text{n}_0$

Where n₀ original frequency of source of sound

v = speed of sound in medium

$\therefore​​​​​​\text{n}'>\text{n}_0\ \ \text{v}_\text{s}=$ speed of source

When source is moving away from observer

$\text{n}'=\frac{\text{v}}{(\text{v}+\text{v}_\text{s})}\text{n}_0\ \text{n}''<\text{n}_0$

Hence, the frequencies in both cases are same and $\text{n}'>\text{n}''.$ so graph (c) verifies the answer.

2. 0.5 second

Explanation:

M = mass string 2.5kg, l = 20m 

M = mas per unit length $=\frac{\text{M}}{\text{l}}=\frac{2.5}{20}=0.125\text{kg/ m}$

$\text{v}=\sqrt{\frac{\text{T}}{\mu}}=\sqrt{\frac{200}{0.125}}=\sqrt{1600}=40\text{m/ s}$

time $= \frac{\text{distance}}{\text{speed}}=\frac{20\text{m}}{40\text{m/ s}}=\frac{1}{2}\sec=0.5\sec.$

4. 1

Explanation:

As fundamental frequency of vibration is:

$\nu=\frac{1}{2\text{l}}\sqrt{\frac{\text{T}}{\text{m}}}$

$\therefore \nu\propto \frac{1}{\text{l}\sqrt{\text{m}}}$ $(\because $ T is constant$)$

$\text{m}=\pi\text{r}^2\rho$

$\therefore \text{m}\propto \text{r}^2$

$\nu=\frac{1}{\text{l}\sqrt{\text{m}}}\propto \frac{1}{\text{lr}}$

$\frac{\nu_1}{\nu_2}=\frac{\text{l}_2\text{r}_2}{\text{l}_1\text{r}_1}$

$=\Big(\frac{2\text{L}}{\text{L}}\Big)\Big(\frac{\text{r}}{2\text{r}}\Big)=1$

2. 4

Explanation:

Beat frequency $=\text{f}_2-\text{f}_1=\frac{\omega_2-\omega_1}{2\pi}$

$=\frac{2008\pi-2000\pi}{2\pi}=4\text{Hz}$