Physics Part Ii Chapter 15 Waves

C.

2

p₁ =p_o sin 2π(x- 1)t
p₂ =p_o sin 2π(x)t
p₃ =p_o sin 2π(x+ 1)t
p=p₁+p₃+p₂
=p_osin2π(x-1)t +p_o sin 2π(x+1)t +p_o sin 2π(x)t
= 2p_osin2πxtcos2πt +p_osin2πxt
= 2p_osin2πxt[2cos πt +1]
⇒ f_beat = 2

A.

α = 25.00 π, β = π

y = 0.005 cos (αx − βt)
comparing the equation with the standard form,

C.

15 ms^-1

C.

196 Hz

|f₁−f₂| =4 Since mass of second tuning fork increases so f₂ decrease and beats increase so f₁>f₂ ⇒ f₂=f₁−4 = 196

D.

20%

B.

5 m/s

As given,
 ........ (i)
Comparing it with

y = a sin(ωt + kx + φ) ............. (ii)
we find ω = 100 rad/sec, k = 20 per metre

B.

20 Hz

Two successive frequencies of closed pipe

C.

ab/(a-b)

According to given figure,

Let the radius of curvature of the common  internal film surface of the double formed by two bubbles be r. Excess of pressure as compared to the atmosphere inside A is (4T/a). In a double bubble, the pressure difference between A and B on either side of the common surface.

$$\begin{gathered} = \frac{4\mathrm{T}}{\mathrm{b}} = \frac{4\mathrm{T}}{\mathrm{a}} \\ \mathrm{This} \mathrm{will} \mathrm{be} \mathrm{equal} \mathrm{to} (4\mathrm{T}/\mathrm{r}) \\ \therefore \frac{4\mathrm{T}}{\mathrm{r}} = \frac{4\mathrm{T}}{\mathrm{b}} -\frac{4\mathrm{T}}{\mathrm{a}} \\ \mathrm{or} \\ \frac{1}{\mathrm{r}} = \left(\frac{1}{\mathrm{b}} - \frac{1}{\mathrm{a}}\right) = \frac{\mathrm{a}-\mathrm{b}}{\mathrm{ba}} \\ \mathrm{or} \mathrm{r} = \frac{\mathrm{ab}}{\mathrm{a}-\mathrm{b}} \end{gathered}$$

B.

αω

Given equation is x = α cos (ωt-θ)

$$\begin{gathered} \mathrm{So} \mathrm{velocity} \left|\frac{\mathrm{dx}}{\mathrm{dt}}\right| =\mathrm{\alpha } \mathrm{\omega sin}\left(\mathrm{\omega }t-\mathrm{\theta }\right) \\ at \left(\mathrm{\omega t}-\mathrm{\theta }\right) ={90}^{\mathrm{o}},\mathrm{the} \mathrm{velocity} \mathrm{will} \mathrm{be} \mathrm{maximum}, \mathrm{therefore} \mathrm{the} \mathrm{maximum} \mathrm{v} \mathrm{velocity} \mathrm{is} \\ \mathrm{\nu } = \alpha \omega \end{gathered}$$

D.

elliptical

$$\begin{gathered} \mathrm{Let} \mathrm{two} \mathrm{perpendicular} \mathrm{simple} \mathrm{harmonic} \mathrm{motion}s \mathrm{are} \\ \mathrm{x} = \mathrm{A} \sin \left(\mathrm{\omega t}\right) ....\left(1\right) \\ \mathrm{y}= B \sin \left(\mathrm{\omega t} +\mathrm{\varphi }\right) ...\left(2\right) \\ \mathrm{from} \mathrm{equation} \left(1\right) \\ \sin \left(\mathrm{\omega t}\right) =\frac{\mathrm{x}}{\mathrm{A}} \\ \Rightarrow \sin { }^2\mathrm{\omega t} + {\cos }^2 \mathrm{\omega t} = 1 \\ \Rightarrow {\cos }^2 \mathrm{\omega t} = 1 - \frac{{\mathrm{x}}^2}{{\mathrm{A}}^2} \\ \Rightarrow \mathrm{cos\omega t} =\sqrt{1-\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2}} \\ \left(2\right)\Rightarrow \frac{\mathrm{y}}{\mathrm{B}} = \sin \mathrm{\omega t} \mathrm{cos\varphi } + \mathrm{cos\omega t} \mathrm{sin\varphi } \\ \frac{\mathrm{y}}{\mathrm{B}}-\frac{\mathrm{x}}{\mathrm{A}}\mathrm{cos\varphi } =\sqrt{1-\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2}} \mathrm{sin\varphi } \\ \mathrm{squaring} \mathrm{on} \mathrm{both} \mathrm{side} \\ \frac{{\mathrm{y}}^2}{{\mathrm{B}}^2} +\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2}{\cos }^2\mathrm{\varphi } -2 \frac{\mathrm{y}}{\mathrm{B}} \frac{\mathrm{x}}{\mathrm{A}}\mathrm{cos\varphi } \\ =\left(1-\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2}\right) {\sin }^2\mathrm{\varphi } \\ \frac{{\mathrm{y}}^2}{{\mathrm{B}}^2} +\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2}\left({\cos }^2\mathrm{\varphi } +{\sin }^2\mathrm{\varphi }\right) \\ -\frac{2\mathrm{xY}}{\mathrm{AB}}\cos \mathrm{\varphi } = {\sin }^2\mathrm{\varphi } \\ \Rightarrow \frac{{\mathrm{y}}^2}{{\mathrm{B}}^2} +\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2} -\frac{2\mathrm{xy}}{\mathrm{AB}} \mathrm{cos\varphi }= {\sin }^2\mathrm{\varphi } \\ \mathrm{When} \mathrm{\varphi } =\frac{\mathrm{\pi }}{2}\Rightarrow \mathrm{cos\varphi } =0 \mathrm{sin\varphi } = 1 \\ \frac{{\mathrm{y}}^2}{{\mathrm{B}}^2}+\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2} = 1 ....\left(3\right) \end{gathered}$$

A.

1:2

The fundamental (first harmonic) for an open-end pipe needs to be an antinode

at both ends, since the air can move at both ends. The lowest frequency is called the 

fundamental frequency or the 1st harmonic. The higher frequencies are called overtones.

Avoiding end correction, the length of the closed organ pipe is

$$\begin{gathered} {\mathrm{l}}_1 =\frac{{\mathrm{\lambda }}_1}{4} \mathrm{or} {\mathrm{\lambda }}_1 =4 {\mathrm{l}}_1 \\ \mathrm{The} \mathrm{length} \mathrm{of} \mathrm{open} \mathrm{organ} \mathrm{pipe} \mathrm{is} \\ {\mathrm{l}}_2=\frac{{\mathrm{\lambda }}_2}{2} \mathrm{or} {\mathrm{\lambda }}_2 =2 {\mathrm{l}}_2 \\ \mathrm{here}, {\mathrm{n}}_1= {\mathrm{n}}_2 \\ \mathrm{so} \frac{\mathrm{v}}{{\mathrm{\lambda }}_1} =\frac{\mathrm{\nu }}{{\mathrm{\lambda }}_2} \mathrm{or} \frac{\mathrm{\nu }}{4 {\mathrm{l}}_1}=\frac{\mathrm{\nu }}{2 {\mathrm{l}}_{ 2}} \\ \mathrm{Therefore}, {\mathrm{l}}_1:{\mathrm{l}}_2 = 1:2 \end{gathered}$$

B.

99 Hz

Each tuning forck gives 3 beats with the next, so the difference in the frequencies of two consecutive forks is 3.

$$\begin{gathered} {\mathrm{f}}_{26} ={\mathrm{f}}_1 +\left(\mathrm{n}-1\right) \times \left(-3\right) \\ \mathrm{f} = 2\mathrm{f} +\left(26 - 1\right)\times \left(-3\right) \\ \mathrm{f} = 75 \mathrm{Hz} \\ \therefore \mathrm{Frequency} \mathrm{of} \mathrm{the} 18 \mathrm{th} \mathrm{tunning} \mathrm{fork} \\ {\mathrm{f}}_{18} ={\mathrm{f}}_1 +\left( 18 - 1\right)\times \left( -3\right) \\ =2 \times 75 + 17 \times \left(-3\right) \\ \mathrm{f} =150 - 51 \\ =99 \mathrm{Hz} \end{gathered}$$

B.

6

Observed frequency of first train

$$\begin{gathered} {\mathrm{n}}_1 =\frac{\mathrm{\nu }}{\mathrm{\nu } - {\mathrm{\nu }}_{\mathrm{s}}}\times \mathrm{n} \\ =\frac{320}{320 - 4}\times 240 \\ =\frac{320}{316} \times 240 \\ =243 \mathrm{Hz} \end{gathered}$$

Observed frequency of second train

$$\begin{gathered} {\text{n}}_2 = \frac{\mathrm{\nu }}{\mathrm{\nu } - {\mathrm{\nu }}_{\mathrm{s}}} \\ = \frac{320}{320 -4}\times 240 \\ = \frac{320}{216}\times 240 \\ = 237 \mathrm{hz} \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{beats} = 243 - 237 \\ \mathrm{Number} \mathrm{of} \mathrm{beats} = 6 \end{gathered}$$

A.

9:4

The average power per unit area that is incident perpendicular to the direction of propagation is called the intensity.

Intensity of sound

$$\begin{gathered} \mathrm{I} = \frac{\mathrm{P}}{4{\mathrm{\pi r}}^2} \\ \Rightarrow \mathrm{i} \propto \frac{1}{{\mathrm{r}}^2} \left(\mathrm{intensity} \propto \frac{1}{{\mathrm{distance}}^2}\right) \\ \Rightarrow \frac{{\mathrm{I}}_1}{{\mathrm{I}}_2} = {\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^2 \\ \mathrm{Here} {\mathrm{r}}_{1 }= 2\mathrm{m}, {\mathrm{r}}_2 = 3\mathrm{m} \end{gathered}$$

Substiting the values, we have

$\frac{{\mathrm{I}}_1}{{\mathrm{I}}_2} ={\left(\frac{3}{2}\right)}^2 =\frac{9}{4}$

B.

180

To reach the solution the given wave equations must be compared with standard equation of progressive wave.

So

y₁ = 4sin 500 Πt              ....(i)

y₂ = 2sin 506 Πt             .....(ii)

Comparing Eqs. (i) and (ii) with
y  = a sin ωt

We have

$$\begin{gathered} {\mathrm{\omega }}_1 = 500 \mathrm{\pi } \\ \Rightarrow {\mathrm{f}}_1 = \frac{500\mathrm{\pi }}{2\mathrm{\pi }} \\ \Rightarrow {\mathrm{f}}_1 = 250 \mathrm{beats}/\mathrm{s} \\ \mathrm{and} {\mathrm{\omega }}_2 = 506 \mathrm{\pi } \\ {\mathrm{f}}_2 = \frac{506\mathrm{\pi }}{2\mathrm{\pi }} \\ {\mathrm{f}}_2 = 253 \mathrm{beats}/\mathrm{s} \\ \mathrm{Thus}, \mathrm{number} \mathrm{of} \mathrm{beats} \mathrm{produced} \\ = {\mathrm{f}}_2 - {\mathrm{f}}_1 \\ = 253 - 250 \\ =3 \mathrm{beats}/\mathrm{s} \\ =3 \times 60 \mathrm{beats}/\min \end{gathered}$$

number of beats = 180 beats/min

D.

20 %

Given:- 

$$\begin{gathered} \Rightarrow {\mathrm{\nu }}_{\mathrm{o}} =\frac{\mathrm{\nu }}{5} \\ \Rightarrow {\mathrm{\nu }}_{\mathrm{o}} = \frac{320}{5} \\ \Rightarrow {\mathrm{\nu }}_{\mathrm{o}} = 64 \mathrm{m}/\mathrm{s} \end{gathered}$$

When observer moves towards the stationary source then

$$\begin{gathered} \mathrm{n}\text{'} = \left(\frac{\mathrm{\nu } + {\mathrm{\nu }}_{\mathrm{o}}}{\mathrm{\nu }}\right) \mathrm{n} \\ \mathrm{n}\text{'} =\left( \frac{320 + 64}{320}\right) \mathrm{n} \\ \mathrm{n}\text{'} = \left(\frac{384}{320}\right) \mathrm{n} \\ \mathrm{n}\text{'} = \frac{384}{320} \end{gathered}$$

Hence percentage increase

$$\begin{gathered} \left(\frac{\mathrm{n}\text{'} - \mathrm{n}}{\mathrm{n}}\right) = \left(\frac{384 - 320}{320} \times 100\right) \\ = \left(\frac{64}{320} \times 100\right)\% \\ \left(\frac{\mathrm{n}\text{'} - \mathrm{n}}{\mathrm{n}}\right) = 20\% \end{gathered}$$

A.

2

Sabine formula for reverberation time is

             $\mathrm{T} = \frac{0.16}{\mathrm{\Sigma } \mathrm{a} \mathrm{S}}$

Where V is volume of hall in m³

∑ $\mathrm{a} \mathrm{S}$ = ${\mathrm{a}}_1{\mathrm{S}}_1 + {\mathrm{a}}_2{\mathrm{S}}_2 +{\mathrm{a}}_3{\mathrm{S}}_3....$

         = total absorption of the hall (room)

Here, S₁, S₂ , S₃, ..... are surface areas of the absorbers and  a_1, a₂ , a₃ .... their respective absorption coefficients 

$$\begin{gathered} \frac{\mathrm{T}\text{'}}{\mathrm{T}} = \frac{\mathrm{V}\text{'}}{\mathrm{S}\text{'}}\times \frac{\mathrm{S}}{\mathrm{V}} \\ = \frac{{\displaystyle {\left(2\right)}^3}}{{\displaystyle {\left(2\right)}^2}} \\ = \frac{8}{4} \end{gathered}$$

$\frac{\mathrm{T}\text{'}}{\mathrm{T}} = 2$

T' = 2T = 2 × 1

T' = 2 s

B.

The sound waves in air are longitudinal while the light waves are transverse

In a longitudinal wave, the particles of the medium oscillate about their mean or equilibrium position along the direction of propagation of the wave itself. Sound waves are longitudinal in nature. In transverse wave, the particles of the medium oscillate about their mean or equilibrium position at right angles to the direction of propagation of wave itself. Light waves being electromagnetic are transverse waves.

D.

415 Hz

When length of air column is $\frac{\mathrm{\lambda }}{4}$, then first resonance occurs.

If we adjust the length of air-column in closed organ pipe as such its any natural frequency equals to the frequency of tuning fork, then the amplitude of forced vibrations of air-column increases very much. This is the state of resonance. At first resonance

                             l = $\frac{\mathrm{\lambda }}{4}$

So, frequency of tuning fork

                         $$\begin{gathered} \mathrm{f} = \frac{\mathrm{v}}{\mathrm{\lambda }} \\ \mathrm{f}= \frac{\mathrm{v}}{4\mathrm{l}} \end{gathered}$$

Given, l= 20 cm= 0.2 m, v = 332 m/s 

Hence 

                        $\mathrm{f} = \frac{332}{4 \times 0.2}$

                        f = 415 Hz

D.

1

Fundamental frequency

           $\mathrm{v} = \frac{1}{2\mathrm{l}}\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$

          $\mathrm{v} = \frac{1}{2\mathrm{l}} \sqrt{\frac{\mathrm{T}}{2{\mathrm{\pi r}}^2 \mathrm{\rho }}}$

Where m = mass per unit length of wire

⇒       $\mathrm{v} \propto \frac{1}{\mathrm{lr}}$

∴      $\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2} = \frac{{\mathrm{l}}_2{\mathrm{r}}_2}{{\mathrm{l}}_1{\mathrm{r}}_1}$

             = $\frac{2\mathrm{L} \times \mathrm{r}}{\mathrm{L} \times 2\mathrm{r}}$

        $\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2} =1$

D.

320 Hz

There are 4 beats between A and B, therefore the possible frequencies of A are 316 or 324 that is (320 ± 4)Hz. 

When the prong of A is filed, its frequency becomes greater than the original frequency. If we assume that original frequency of A is 324 then on filing its frequency will be greater than 324. The beats between A and B will be more than 4. But it is given that the beats are again 4, therefore, 324 is not possible. 

Therefore, required frequency must be 316 Hz. (This is true, because on filing the frequency may increase so as to give 4 beats with B of frequency 320 Hz)

B.

The sound waves in air are longitudinal while the light waves are transverse

Sound waves are longitudinal waves, meaning that the waves propagate by compression and rarefaction of their medium. They are termed longitudinal waves because the particles in the medium through which the wave travels oscillate parallel in direction of motion. 

Longitudinal waves transmit energy by compressing and refracting the medium in the same direction as they are travelling. Sound waves are longitudinal waves are travel by compressing the air through which they travel, causing vibration.

Light and sound waves both are forms of wave motion, sound requires a solid, liquid or gaseous medium; whereas light travels through empty space. the denser the medium, the greater the speed of sound.

D.

415 Hz

If we adjust the length of air-column in closed organ pipe as such its any natural frequency equals to the frequency of tuning fork, then the amplitude of forced vibrations of air-column increases very much. This is the state of resonance.

At first resonance

              $\mathrm{l} =\frac{\mathrm{\lambda }}{4}$

So, frequency of tuning fork

            f = $\frac{\mathrm{v}}{\mathrm{\lambda }}$

            f = $\frac{\mathrm{v}}{4 \mathrm{l}}$

Given, l = 20 cm = 0.2 m

          v = 332 m/s

Hence

          $\mathrm{f} = \frac{0.332}{4 \times 0.2}$

         f = 415 Hz

A.

${\mathrm{y}}_{\mathrm{o}} \frac{\mathrm{\pi }}{2}$

Given equation is

       $\mathrm{y} = {\mathrm{y}}_{\mathrm{o}} \sin 2\mathrm{\pi f} \mathrm{t}$

Where $\mathrm{A} = {\mathrm{y}}_{\mathrm{o}} , \mathrm{k} = \frac{2\mathrm{\pi }}{\mathrm{\lambda }}$

According to the question,

          v = 4 ${\mathrm{v}}_{\mathrm{p}}$

∴         Aω = $4 \frac{\mathrm{\omega }}{\mathrm{k}}$

∴        ${\mathrm{y}}_{\mathrm{o}} \mathrm{\omega } =\frac{ 4 \mathrm{\omega }}{\mathrm{k}}$

⇒        ${\mathrm{y}}_{\mathrm{o}} = \frac{4}{\mathrm{k}}$

⇒        ${\mathrm{y}}_{\mathrm{o}} = \frac{4}{2\mathrm{\pi }} \times \mathrm{\lambda }$

 Therefore, wavelength $\mathrm{\lambda } = \frac{\mathrm{\pi } {\mathrm{y}}_{\mathrm{o}}}{2}$

B.

Frequency

Sound propagated in the form of longitudinal waves. When sound travels from one medium to another, both its velocity and wavelength undergo change. The velocity of sound in a given is obtained by equation 

          v = nλ

Where n - frequency of sound and λ is the wavelength of the medium. The velocity of sound is directly proportional to the wavelength.

The frequency of sound depends upon the source of the sound, not the medium of propagation. Hence, it does not change. 

C.

they are heavily absorbed by the atmosphere

The electric waves do not have the capacity to travel through the atmosphere. They get immediately grounded and they become neutral as fast as possible. This makes electrical waves incapable to travel through the atmosphere.

C.

1.6 KHz

The reflected sound appears to propagate in a direction opposite to that of moving engine. Thus, the source and the observer can be presumed to approach each other with same velocity. 

The Doppler effect is observed whenever the source of waves is moving with respect to an observer. 

So according to Doppler effect

      v' = $\frac{\mathrm{v} ( \mathrm{v} + {\mathrm{v}}_{\mathrm{o}})}{\left(\mathrm{v} - {\mathrm{v}}_{\mathrm{s}}\right)}$

                = v $\frac{\mathrm{v} + {\mathrm{v}}_{\mathrm{s}}}{\mathrm{v} - {\mathrm{v}}_{\mathrm{s}}}$                        ($\because {\mathrm{v}}_{\mathrm{o}} = {\mathrm{v}}_{\mathrm{s}}$)

⇒         v' = 1.2 $\left(\frac{350 + 50}{350 - 50}\right)$

                = $\frac{1.2 \times 400}{300}$

           v' = 1.6 kHz

B.

10 m

The given equation of a progressive wave is

         y = 3 sin $\mathrm{\pi } \left(\frac{\mathrm{t}}{2}- \frac{\mathrm{x}}{4}\right)$

            = 3 sin 2 $\mathrm{\pi }\left(\frac{\mathrm{t}}{4 } - \frac{\mathrm{x}}{8}\right)$

The standard equation of a progressive wave is

        y = y_o sin $2\mathrm{\pi } \left(\frac{\mathrm{t}}{\mathrm{T}} - \frac{\mathrm{x}}{\mathrm{\lambda }}\right)$

Comparing these two equations, we get

           T = 4 s ,   λ = 8m

∴  The velocity of wave,

    v = $\frac{\mathrm{\lambda }}{\mathrm{T}}$

      = $\frac{8}{4}$

⇒   v = 2 ms^-1

Distance travelled by wave in time t is

       s = v t

⇒     s = 2 × 5

⇒        s = 10 m

A.

$\mathrm{\lambda } = \mathrm{\pi } {\mathrm{y}}_{\mathrm{o}}$

A transeverse wave is described by equation

                y = ${\mathrm{y}}_{\mathrm{o}} \sin 2\frac{\mathrm{\pi } }{\mathrm{\lambda }}\left( \mathrm{vt} - \mathrm{k} \right)$

Comparing the given equation of the standard equation of wave

         w y_o = 2 v

           2πn y_o = 2ny

          λ = π y_o

C.

remain same

organ pipe - Sound producing such a tube which is close at one end and open at other end is called closed organ pipe, and the tube which open at both the ends called open organ pipe.

Frequency of open organ pipe is double than that of closed pipe.

If frequency of open pipe =  n

∴  Frequency of closed pipe = $\frac{\mathrm{n}}{2}$

On inserting an open pipe into water upto half of its height, it will behave like a closed pipe of half length and frequency is inversely proportional to the length of air column.

Hence, frequency of closed pipe will become double

                                      = 2 × $\frac{\mathrm{n}}{2}$

                                       = n

B.

273 Hz

An increase (or decrease ) in the frequency of sound, light or other waves as the source and observer move towards (or away from) other. The effect causes the sudden change in pitch noticeable in a passing siren, as well as the redshift seen by astronomers. 

Here source is moving and the observer is stationary so according to Doppler effect we use the following relation.

Apparent frequency n' = n $\left(\frac{\mathrm{V}}{\mathrm{V} + {\mathrm{V}}_{\mathrm{s}}}\right)$

  V = Velocity of sound

  V_s = velocity of source

  ∴   n' = $300 \left(\frac{332}{332 + 30}\right)$

           = 273 Hz

A.

0.12 N

Mass per unit length m = 1.3 × 10^-4 kg/m

                               y = 0.021 sin (x + 30t)

Comparing with y = a sin $\left(\mathrm{\omega t} + \frac{2\mathrm{\pi } \mathrm{x}}{\mathrm{\lambda }}\right)$

   a = 0.021,  ω = 30

    $2\mathrm{\pi n} = 30$

⇒    n = $\frac{30}{2\mathrm{\pi }}$

We know that 

     $\frac{2\mathrm{\pi }}{\mathrm{\lambda }} = 1$

∴     $\mathrm{\lambda } = 2\mathrm{\pi }$

Frequency,

       n = $\frac{1}{\mathrm{\lambda }}\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$

      $\frac{30}{2\mathrm{\pi }} = \frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{T}}{1.3 \times {10}^{-4}}}$

       T = 1.3 × 10^-4 × 900

        T = 0.117 

        T = 0.12 N

C.

c + v

According to the Doppler effect 

Here both source and observer moving

Observer and light are moving towards each other, so relative velocity is c + v.

D.

50 m/s

Given:-

m = 19.2 x 10^-3 kg /m

Now see the free body diagram

The equation for the motion

   T - 4g - 4a = 0

⇒  T = 4 ( a + g ) 

        = 4 ( 2 + 10)

      T = 48 N

Now wave speed

    v = $\sqrt{\left(\frac{\mathrm{T}}{\mathrm{m}}\right)}$

      = $\sqrt{2.5 \times {10}^{-3}}$

  v = 50 m/s

C.

$2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

Let when m is hanging, the extension in the spring is l, then tension in the string is,

    T₁ = k l = mg

When force is applied, let the further extension is x.

 The tension in the spring is

     T₂ = k ( x + l )

∴ Driving force = T₂ - T₁

                       = k (  x + l ) - kl

  Driving force = k x

∴   Acceleration  = $\frac{\mathrm{kl}}{\mathrm{m}}$

⇒          T = $2\mathrm{\pi } \sqrt{\frac{\mathrm{displacement}}{\mathrm{acceleration}}}$

               = $2\mathrm{\pi } \sqrt{\frac{\mathrm{x}}{\mathrm{kx}}}$

⇒         T = $2\mathrm{\pi } \sqrt{\frac{\mathrm{x}}{{\displaystyle \frac{\mathrm{kx}}{\mathrm{m}}}}}$

⇒         T = $2\mathrm{\pi } \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

B.

4.34 x 10⁴ Hz, 4.67 x 10⁴ Hz

D.

4.67 x 10³ Hz, 4.34 x 10⁴ Hz

Frequency received by bat after reflection from wall (1),

      f₁ = f $\left[\frac{\mathrm{v} + \mathrm{u}}{\mathrm{v} - \mathrm{u}}\right]$

          = 4.5 × 10⁴ $\left[ \frac{330 + 6}{330 - 6} \right]$

                          [ given f = 4.5 × 10⁴ Hz ]

         = 4.67 × 10⁴ Hz

Frequency received by bat after reflection from wall (2),

     f₂ =  f $\left[ \frac{\mathrm{v} - \mathrm{u}}{\mathrm{v} + \mathrm{u}} \right]$

         = 4.5 × 10⁴ $\left[ \frac{330 - 6}{330 + 6} \right]$

      f₂ = 4.34 × 10⁴ Hz

C.

0.6 × 10³ m/s

Phase velocity:- It is the ratio of the angular frequency to the wave number.

                v_p = $\frac{\mathrm{\omega }}{\mathrm{k}}$

k = wave number

ω = angular frequency

             $\frac{{\mathrm{dv}}_{\mathrm{p}}}{{\mathrm{v}}_{\mathrm{p}}} = \frac{\mathrm{d\omega }}{\mathrm{\omega }} - \frac{\mathrm{dk}}{\mathrm{k}}$

∴            $\frac{\mathrm{dk}}{\mathrm{k}} = \frac{\mathrm{d\omega }}{\mathrm{\omega }} - \frac{{\mathrm{dv}}_{\mathrm{p}}}{{\mathrm{v}}_{\mathrm{p}}}$

                     = 2 % $- \left(- 3 \%\right)$

                $\frac{\mathrm{dk}}{\mathrm{k}}$ = 5%

Group velocity:- it is the derivative of the angular frequency with respect to the wave number.

Group velocity is given by

                v_g = $\frac{\mathrm{d\omega }}{\mathrm{dk}}$

              $\frac{\mathrm{d\omega }}{\mathrm{dk}} = {\mathrm{v}}_{\mathrm{p}} \left(\frac{{\displaystyle \raisebox{1ex}{$\mathrm{d\omega }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{\omega }$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{dk}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{k}$}\right.}}\right)$

                     = 1.5 × 10³ $\left(\frac{2}{5}\right)$

               $\frac{\mathrm{d\omega }}{\mathrm{dk}}$ = 0.6 × 10³ m/s

D.

195 Hz

Tunning fork are sounded together 5 beats per second are produced or 5Hz

Let frequency of fork = n

Then probably frequencies of tunning fork due to beats due to beats formation are 

     ( n + 5 ) Hz and ( n $-$ 5 ) Hz

We know

Frequency is inversely proportional to length of wire

∴    $\frac{\mathrm{n} + 5}{\mathrm{n} - 5} = \frac{100}{95}$

⇒  95 ( n + 5 ) = 100 ( n $-$ 5 )

⇒  19 ( n + 5 ) = 20 ( n $-$ 5 )

⇒   19n + 95 = 20n $-$ 100

⇒  95 + 100 = 20n $-$ 19n

⇒            n = 195

Hence frequency of the fork is 195 Hz

B.

$\frac{1}{2\mathrm{\pi }}\sqrt{\frac{4\mathrm{k}}{\mathrm{M}}}$

Two springs on the L.H.S. of mass M are in series and two springs on the R.H.S. of mass M are in parallel. These combinations of springs will be considered in parallel to mass M. Thus effective spring constant,

     K = $\frac{2\mathrm{k} \times 2\mathrm{k}}{2\mathrm{k} + 2\mathrm{k}}$ + ( k + 2k)

        = 4k

∴ Frequency ν = $\frac{1}{2\mathrm{\pi }} \sqrt{\frac{\mathrm{K}}{\mathrm{M}}}$

                   ν  = $\frac{1}{2\mathrm{\pi }} \sqrt{\frac{4\mathrm{k}}{\mathrm{M}}}$

D.

All are correct

 The equation of a given progressive wave is

         y = 5 sin $\left( 100 \mathrm{\pi t} - 0.4 \mathrm{\pi x} \right)$            ....(i)

The standard equation of a progressive wave is

        y = a sin ( ωt $-$ kx )

Comparing equation (i) and (ii)

     α = 5 m

     ω = 100 $\mathrm{\pi }$ rad s^-1

      k = 0.4 $\mathrm{\pi } {\mathrm{m}}^{-1}$

(1) Amplitude of the wave, a = 5 m

(2) Wavelength of the wave,

         $\mathrm{\lambda } = \frac{2\mathrm{\pi }}{\mathrm{k}}$

            = $\frac{2 \mathrm{\pi }}{0.4 \mathrm{\pi }}$

         λ = 50 m

(3) Frequency of he wave

        $v= \frac{\mathrm{\omega }}{2 \mathrm{\pi }}$

          = $\frac{100\mathrm{\pi }}{2 \mathrm{\pi }}$

      v = 50 Hz

(4) Velocity of the wave

       v = v λ

           = ( 50 s^-1 ) (5 m)

       v = 250 ms^-1

A.

If both assertion and reason are true and reason is the correct explanation of assertion.

The fundamental frequency of open organ pipe is 

                  n = $\frac{\mathrm{v}}{2 \mathrm{l}}$

As temperature increase, both v and I increase but I increase more rapidly than I. Hence the fundamental frequency increases as the temperature increases.