Physics 5 Marks Questions

$\text{u}=334\text{m/s}, \text{v}_\text{b}=4\sqrt{2}\text{m/s},\ \text{v}_0=0$

So, $\text{v}_\text{s}=\text{V}_\text{b}\cos\theta=4\sqrt{2}\times\Big(\frac{1}{\sqrt{2}}\Big)=4\text{m/s}$

So, the apparent frequency $\text{f}'=\Big(\frac{\text{u}+0}{\text{u}-\text{v}_\text{b}\cos\theta}\Big)\text{f}=\Big(\frac{334}{334-4}\Big)\times1650=1670\text{Hz}.$

Here given $\lambda=\frac{\text{d}}{2}$

Initial path difference is given by $2\sqrt{\Big(\frac{\text{d}}{2}\Big)^2+2\text{d}^2}-\text{d}$

If it is now shifted a distance x then path difference will be

$=2\sqrt{\Big(\frac{\text{d}}{2}\Big)^2}+\Big(\sqrt{2}\text{d}+\text{x}\Big)^2-\text{d}=\frac{\text{d}}{4}\Big(2\text{d}+\frac{\text{d}}{4}\Big)$

$\Rightarrow\Big(\frac{\text{d}}{2}\Big)^2+\Big(\sqrt{2}\text{d}+\text{x}\Big)^2=\frac{169\text{d}^2}{64}$

$\Rightarrow\frac{153}{64}\text{d}^2$

$\Rightarrow\sqrt{2}\text{d}+\text{x}=1.54\text{d}$

$\Rightarrow\text{x}=1.54\text{d}-1.414\text{d}$

$=0.13\text{d}.$

1. Given that, f = 1200Hz, u = 170m/s, L = 200m, v = 340m/s

From Doppler’s equation (as in problem no.84)

$\text{f}'=\text{f}\Big(\frac{\text{v}^2}{\text{v}^2-\text{u}^2}\Big)=1200\times\frac{340^2}{340^2-170^2}=1600\text{Hz}.$

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2. v = velocity of sound, u = velocity of source let, t be the time taken by the sound to reach at D

$\text{DO}=\text{vt}'=\text{L},$ and $\text{S}'\text{O}=\text{ut}'$

$\text{t}'=\frac{\text{L}}{\text{V}}$

$\text{S}'\text{D}=\sqrt{\text{S}'\text{O}^2+\text{DO}^2}=\sqrt{\text{u}^2\frac{\text{L}^2}{\text{v}^2}+\text{L}^2}=\frac{\text{L}}{\text{v}}\sqrt{\text{u}^2+\text{v}^2}$

Putting the values in the above equation, we get

$\text{S}'\text{D}=\frac{220}{340}\sqrt{170^2+340^2}=223.6\text{m}.$

1. The frequency by the passenger sitting near the open window is 500Hz. The speed of sound in air V = 340m/ s.

Speed of the source u ₹ = 180km/ h

$=\frac{108000}{3600}\text{m}/\text{s}=30\text{m}/\text{s}.$

1. After the train has passed the apparent frequency heard by a person standing near the track will be,

So, $\text{f}''=\Big(\frac{340+0}{340+30}\Big)\times500=459\text{Hz}$

3. The person inside the source will listen the original frequency of the train.

Here, given $\text{V}_\text{m}=10\text{m/s}$

For the person standing near the track

Apparent frequency $=\frac{\text{u}+\text{V}_\text{m}+0}{\text{u}+\text{V}_\text{m}-(-\text{V}_\text{s})}\times500=458\text{Hz}.$

1. Here given $\text{n}=100,\ \text{v}=350\text{m/s}$

$\Rightarrow\lambda=\frac{\text{v}}{\text{n}}$

$=\frac{350}{100}=3.5\text{m}$

In 2.5ms, the distance travelled by the particle is given by

$\triangle\text{x}=350\times2.5\times10^{-3}$

So, phase difference

$\phi=\frac{2\pi}{\lambda}\times\triangle\text{x}\Rightarrow\frac{2\pi}{\Big(\frac{350}{100}\Big)}\times350\times2.5\times10^{-3}=\Big(\frac{\pi}{2}\Big).$

2. In the second case, Given $\triangle\eta=10\text{cm}=10^{-1}\text{m}$

So, $\phi=\frac{2\pi}{\text{x}}\triangle\text{x}=\frac{2\pi\times10^{-1}}{\Big(\frac{350}{100}\Big)}=\frac{2\pi}{35}.$

$\text{f}=2\text{KHz}, \text{v}=330\text{m/s},\ \text{u}=22\text{m/s}$

At t = 0, the source crosses P

1. Time taken to reach at Q is

$\text{t}=\frac{\text{S}}{\text{v}}=\frac{330}{330}=1\text{sec}$

2. The frequency heard by the listner is

$\text{f}'=\text{f}\Big(\frac{\text{v}}{\text{v}-\text{u}\cos\theta}\Big)$

since, $\theta=90^\circ$

$\text{f}'=2\times\Big(\frac{\text{v}}{\text{u}}\Big)=2\text{KHz}.$

3. After 1sec, the source is at 22m from P towards right.

According to the questions velocity of car A = V_A = 108km/h = 30m/s

V_B = 72km/h = 20m/s, f = 800Hz

So, the apparent frequency heard by the car B is given by,

$\text{f}'\Big(\frac{330-20}{330-30}\Big)\times800$

$\Rightarrow826.9=827\text{Hz}.$

Given that, r = 0.17m, F = 800Hz, u = 340m/s

Frequency band = f₁ - f₂ = 6Hz

Where f₁ and f₂ correspond to the maximum and minimum apparent frequencies (both will occur at the mean position because the velocity is maximum).

Now, $\text{f}_1\Big(\frac{340}{340+\text{v}_\text{s}}\Big)\text{f}$ and $\text{f}_2=\Big(\frac{340}{340+\text{v}_\text{s}}\Big)\text{f}$

$\therefore\text{f}_1-\text{f}_2=8$

$\Rightarrow340\text{f}\Big(\frac{1}{340-\text{v}_\text{s}}-\frac{1}{340+\text{v}_\text{s}}\Big)=8$

$\Rightarrow\frac{2\text{v}_\text{s}}{340^2-\text{v}_\text{s}^2}=\frac{8}{340\times800}$

$\Rightarrow340^2-\text{v}_\text{s}^2=68000\text{v}_\text{s}$

Solving for vs we get, $\text{v}_\text{s}=1.695\text{m/s}$

For SHM, $\text{v}_\text{s}=\text{r}\omega$

$\Rightarrow\omega=\Big(\frac{1.695}{0.17}\Big)=10$

So, $\text{T}=\frac{2\pi}{\omega}=\frac{\pi}{5}$

$=0.63\text{sec}.$

$\text{f}=\frac{\text{v}}{4\text{L}_1}$

$440=\frac{330}{4\text{L}_1}$

$\Rightarrow \text{L}_1=\frac{330}{440\times 4}=0.1875\text{m}$

$=18.8\text{cm}$

$\text{f}=\frac{3\text{v}}{4\text{L}_2}$

$\Rightarrow 440=\frac{3\times 330}{4\text{L}_2}$

$\Rightarrow \text{L}_2=\frac{3\times 330}{440\times4}=05.63\text{m}$

$=56.3\text{cm}$

$\triangle\text{L}=6.4-6.0=0.4\text{m}$

Let, the piston resonates at length l₁ and l₂

Here, $\text{l}=32;\ \text{v}=?,\ \text{n}=512\text{Hz}$

Now $\Rightarrow512=\frac{\text{v}}{\lambda}$

$\Rightarrow\text{v}=512\times0.64=328\text{m/s}$

$\text{f}=400\text{Hz},\ \text{u}=324\text{m/s}$

$\text{f}'=\frac{\text{u}-(-\text{v})}{\text{u}-(0)}\text{f}=\frac{324+\text{v}}{324}\times400\ \dots(1)$

for the reflected wave, $\text{f}'=410=\frac{\text{u}-0}{\text{u}-\text{v}}\text{f}'$

$\Rightarrow410=\frac{324}{324-\text{v}}\times\frac{324+\text{v}}{324}\times400$

$\Rightarrow810\text{v}=324\times10$

$\Rightarrow\text{v}=\frac{324\times10}{810}=4\text{m/s}.$

1. Here given $\text{V}_\text{air}=340\text{m/s},\ \text{Power}=\frac{\text{E}}{\text{t}}=20\text{W}$

$\text{f}=2,000\text{Hz},\ \rho=1.2\text{kg/m}^3$

So, intensity $\text{I}=\frac{\text{E}}{\text{t}.\text{A}}$

$=\frac{20}{4\pi\text{r}^2}=\frac{20}{4\times\pi\times6^2}=44\text{mw/m}^2$ (because r = 6m)

2. We know that $\text{I}=\frac{\text{P}_0^2}{2\rho\text{V}_\text{air}}$

$\Rightarrow\text{p}_0=\sqrt{1\times2\rho\text{V}_\text{air}}$

$=\sqrt{2\times1.2\times340\times44\times10^{-3}}$

$=6.0\text{N/m}^2.$

3. We know that $\text{I}=2\pi^2\text{S}_0^2\text{v}^2\rho\text{v}$ where S₀ = displacement amplitude

$\Rightarrow\text{S}_0=\sqrt{\frac{\text{I}}{\pi^2\rho^2\rho\text{V}_\text{air}}}$

Putting the value we get $\text{S}_\text{g}=1.2\times10^{-6}\text{m}.$

1. At A, velocity of the particle is given by,

$\text{v}_\text{A}=\sqrt{\text{rg}}=\sqrt{1.6\times10}=4\text{m/s}$

and at C, $\text{v}_\text{A}=\sqrt{\text{rg}}=\sqrt{1.6\times10}=4\text{m/s}$

So, maximum frequency at C,

$\text{f}'_\text{c}=\frac{\text{u}}{\text{u}-\text{v}_\text{s}}\text{f}=\frac{330}{330-8.9}\times500=513.85\text{Hz}.$

Similarly, maximum frequency at A is given by $\text{f}'_\text{A}=\frac{\text{u}}{\text{u}-(-\text{v}_\text{s})}\text{f}=\frac{330}{330+4}(500)=494\text{Hz}.$

2. Velocity at $\text{B}=\sqrt{3\text{rg}}=\sqrt{3\times1.6\times10}=6.92\text{m/s}$

So, frequency at B is given by, $\text{f}'_\text{B}=\frac{\text{u}}{\text{u}+\text{v}_\text{s}}\times\text{f}=\frac{330}{330+6.92}\times500=490\text{Hz}$

and frequency at D is given by,

$\text{f}_\text{D}=\frac{\text{u}}{\text{u}-\text{v}_\text{s}}\times\text{f}$

$=\frac{330}{330-6.92}\times500$

$=511\text{Hz}$

1. $\text{V}_\text{s}=12\text{km/h}=\frac{10}{3}=\text{m/s}$

$\text{V}_0=0,\ \text{u}=330\text{m/s}$

So, the apparent frequency is given by $\text{f}'=\bigg(\frac{330}{\frac{330-10}{3}}\bigg)\times1600=1616\text{Hz}$

2. The reflected sound from the wall whistles now act as a sources whose frequency is 1616Hz.

So, $\text{u}=330\text{m/s},\ \text{V}_\text{s}=0,\ \text{V}_0=\frac{10}{3}\text{m/s}$

So, the frequency by the man from the wall,

$\Rightarrow\text{f}''=\bigg(\frac{330+\frac{10}{3}}{330}\bigg)\times1616=1632\text{m/s}.$

$\therefore\text{OQ}=\text{ut}$ and $\text{QP}=\text{vt}$

$\cos\theta=\frac{\text{u}}{\text{v}}$

$\text{f}'=\text{f}\Big(\frac{\text{v}-0}{\text{v}-\text{u}\cos\theta}\Big)=\text{f}\bigg(\frac{\text{v}}{\text{v}-\frac{\text{u}^2}{\text{v}}}\bigg)=\text{f}\Big(\frac{\text{v}^2}{\text{v}^2-\text{u}^2}\Big)$