Waves — Page 7 | NEET Physics

Q61

Two stationary sources each emitting waves of wavelength

\lambda

, an observer moves from one source to another with velovcity u. Then number of beats heard by him

A

{{2u} \over \lambda }

B

{u \over \lambda }

C

\sqrt {u\lambda }

D

{u \over {2\lambda }}

Correct Answer

Option A

Solution

f' = {{v - u} \over v}f;f'' = {{v + u} \over v}f

Number of beats =

f'' - f' = {{2u} \over \lambda }

Q62

Two harmonic waves moving in the same direction superimpose to form a wave

x=\mathrm{a} \cos (1.5 \mathrm{t}) \cos (50.5 \mathrm{t})

where t is in seconds. Find the period with which they beat. (close to nearest integer)

A 1 s

B 4 s

C 2 s

D 6 s

Correct Answer

Option C

Solution

The superposition of the two harmonic waves results in the wave equation: $x = a \cos(1.5t) \cos(50.5t)$ This equation can be expanded using the trigonometric identity for the product of cosines: $x = \dfrac{a}{2} \cos[(1.5 + 50.5)t] + \dfrac{a}{2} \cos[(50.5 - 1.5)t]$ Simplifying, we find: $x = \dfrac{a}{2} \cos(52t) + \dfrac{a}{2} \cos(49t)$ This represents two waves with angular frequencies $52$ and $49$ , respectively.

To find the beat frequency, we calculate the differences in frequencies: $f_1 = \dfrac{52}{2\pi}, \quad f_2 = \dfrac{49}{2\pi}$ The beat frequency is then: $f_{\text{Beat}} = f_1 - f_2 = \dfrac{3}{2\pi} \text{ Hz}$ The period of the beats, $T_{\text{Beat}}$ , is the reciprocal of the beat frequency: $T_{\text{Beat}} = \dfrac{1}{f_{\text{Beat}}} = \dfrac{2\pi}{3} \text{ sec}$ Approximating this gives: $T_{\text{Beat}} \approx 2.09 \text{ sec} \approx 2 \text{ sec}$

Q63

Displacement of a wave is expressed as

x(t)=5 \cos \left(628 t+\dfrac{\pi}{2}\right) \mathrm{m}

. The wavelength of the wave when its velocity is

300 \mathrm{~m} / \mathrm{s}

is :

(\pi=3.14)

A 0.33 m

B 0.5 m

C 3 m

D 5 m

Correct Answer

Option C

Solution

\begin{aligned} & \mathrm{x}(\mathrm{t})=5 \cos \left[628 \mathrm{t}+\frac{\pi}{2}\right] \mathrm{m} \\ & \text{ velocity }\left(\mathrm{v}_\omega\right)=300 \mathrm{~m} / \mathrm{s} \\ & \mathrm{v}_{\mathrm{w}}=\frac{\omega}{\mathrm{K}} \\ & 300=\frac{628}{\mathrm{~K}} \Rightarrow \mathrm{~K}=\frac{628}{300} \\ & \frac{2 \pi}{\lambda}=\frac{628}{300} \Rightarrow \lambda=\frac{2 \times 3.14 \times 300}{628} \\ & \lambda=2 \mathrm{~m} \end{aligned}

Q64

In the resonance experiment, two air columns (closed at one end) of 100 cm and 120 cm long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is:

A

370 \mathrm{~m} / \mathrm{s}

B

340 \mathrm{~m} / \mathrm{s}

C

335 \mathrm{~m} / \mathrm{s}

D

360 \mathrm{~m} / \mathrm{s}

Correct Answer

Option D

Solution

The fundamental frequency for a closed (organ) pipe can be expressed as: $f = \dfrac{v}{4\ell}$ For the first air column, with length $\ell_1$ , the frequency $f_1$ is: $f_1 = \dfrac{v}{4\ell_1}$ For the second air column, with length $\ell_2$ , the frequency $f_2$ is: $f_2 = \dfrac{v}{4\ell_2}$ The beat frequency, which is the difference in these two frequencies ( $f_1 - f_2$ ), is given as 15 beats per second: $\text{Beat} = f_1 - f_2 = \dfrac{v}{4} \left( \dfrac{1}{\ell_1} - \dfrac{1}{\ell_2} \right)$ Substitute the given lengths into the formula: $15 = \dfrac{v}{4} \left( \dfrac{1}{1} - \dfrac{1}{1.2} \right)$ Simplify the equation: $15 = \dfrac{v}{4} \left( \dfrac{0.2}{1.2} \right)$ Solve for $v$ : $v = \dfrac{15 \times 4 \times 1.2}{0.2} = 60 \times 6 = 360 \, \text{m/s}$ Thus, the velocity of sound in the air column is 360 m/s.

Q65

A student is performing the experiment of resonance column. The diameter of the column tube is 6 cm. The frequency of the tuning fork is 504 Hz. Speed of the sound at the given temperature is 336 m/s. The zero of the metre scale coincides with the top end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is :

A 13 cm

B 18.4 cm

C 16.6 cm

D 14.8 cm

Correct Answer

Option D

Solution

$\therefore$

l + 1.8 = {\lambda \over 4}

Also

\lambda = {v \over f} = {{336} \over {504}}

\Rightarrow l + 1.8 = {{336} \over {4 \times 504}}

\Rightarrow l = 14.86

cm

Q66

A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is Y = 0.3 sin(0.157x) cos(200pt). The length of the string is : (All quantities are in SI units.)

A 60 m

B 20 m

C 80 m

D 40 m

Correct Answer

Option C

Solution

4th harmonic

4{\lambda \over 2} = l;2\lambda = l

From equation

{{2\pi } \over \lambda } = 0.157

$\lambda$ = 40 ;

l

= 2 $\lambda$ = 80 m

Q67

A transverse wave is represented by

y=2 \sin (\omega t-k x)\, \mathrm{cm}

. The value of wavelength (in

\mathrm{cm}

) for which the wave velocity becomes equal to the maximum particle velocity, will be :

A 4

\pi

B 2

\pi

C

\pi

D 2

Correct Answer

Option A

Solution

{\omega \over k} = A\omega

\Rightarrow k = {1 \over A} = {1 \over {2\,cm}}

\Rightarrow {{2\pi } \over \lambda } = {1 \over {2\,cm}}

\Rightarrow \lambda = 4\pi \,cm

Q68

In the wave equation

y=0.5 \sin \dfrac{2 \pi}{\lambda}(400 \mathrm{t}-x) \,\mathrm{m}

the velocity of the wave will be:

A 200 m/s

B 200

\sqrt2

m/s

C 400 m/s

D 400

\sqrt2

m/s

Correct Answer

Option C

Solution

{v_{wave}} = \left| {{{coefficient\,of\,t} \over {coefficient\,of\,x}}} \right|

= {{400} \over 1} = 400

m/s

Q69

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is

60 \mathrm{~cm}

, the length of the closed pipe will be:

A 15 cm

B 60 cm

C 45 cm

D 30 cm

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{f}_1=\frac{\mathrm{v}}{4 \mathrm{~L}_1} \\ & \mathrm{f}_1=\mathrm{f}_2 \\ & \frac{\mathrm{v}}{4 \mathrm{~L}_1}=\frac{\mathrm{v}}{\mathrm{L}_2} \\ & \Rightarrow \mathrm{L}_2=4 \mathrm{~L}_1 \\ & 60=4 \times \mathrm{L}_1 \\ & \mathrm{~L}_1=15 \mathrm{~cm} \end{aligned}

Q70

In an experiment with a closed organ pipe, it is filled with water by

\left(\dfrac{1}{5}\right)

th of its volume. The frequency of the fundamental note will change by

A

20 \%

B

25 \%

C

-20 \%

D

-25 \%

Correct Answer

Option B

Solution

\begin{aligned} &\lambda_1=4 \ell\\ &\mathrm{f}_1=\frac{\mathrm{v}}{4 \ell} \end{aligned}

\begin{aligned} & \lambda_2=\frac{16 \ell}{5} \\ & \mathrm{f}_2=\frac{5 \mathrm{~V}}{16 \ell} \\ & \frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{\frac{\mathrm{V}}{\ell}\left(\frac{1}{16}\right)}{\frac{\mathrm{V}}{4 \ell}} \times 100=25 \% \end{aligned}