Waves — Page 8 | NEET Physics

Q71

A wave travelling along the

x

-axis is described by the equation

y(x, t)=0.005

\cos \,\left( {\alpha \,x - \beta t} \right).

If the wavelength and the time period of the wave are

0.08

m

and

2.0s

, respectively, then

\alpha

and

\beta

in appropriate units are

A

\alpha = 25.00\pi ,\,\beta = \pi

B

\alpha = {{0.08} \over \pi },\,\beta = {{2.0} \over \pi }

C

\alpha = {{0.04} \over \pi },\,\beta = {{1.0} \over \pi }

D

\alpha = 12.50\pi ,\,\beta = {\pi \over {2.0}}

Correct Answer

Option A

Solution

y\left( {x,t} \right) = 0.005\,\cos \left( {\alpha x - \beta t} \right)

(Given) Comparing it with the standard equation of wave

y\left( {x,t} \right) = a\cos \left( {kx - \omega t} \right)

we get

k = \alpha

\,\,\,\,\,

and

\,\,\,\,\,

\omega = \beta

$\therefore$

{{2\pi } \over \gamma } = \alpha

\,\,\,\,\,

and

\,\,\,\,\,

{{2\pi } \over T} = \beta

$\therefore$

\alpha = {{2\pi } \over {0.08}} = 25\pi

\,\,\,\,\,

and

\,\,\,\,\,

\beta = {{2\pi } \over 2} = \pi

Q72

The equation of a wave travelling on a string is y = sin[20πx + 10πt], where x and t are distance and time in SI units. The minimum distance between two points having the same oscillating speed is :

A 10 cm

B 2.5 cm

C 20 cm

D 5.0 cm

Correct Answer

Option D

Solution

The equation for a wave traveling on a string is given by $y = \sin(20\pi x + 10\pi t)$ , where $x$ is the distance and $t$ is the time in SI units.

To find the minimum distance between two points having the same oscillating speed, we use the concept that this distance is half the wavelength $\left(\dfrac{\lambda}{2}\right)$ .

To find the wavelength $\lambda$ : $\lambda = \dfrac{2\pi}{k}$ Here, the wave number $k = 20\pi$ .

Thus, $\lambda = \dfrac{2\pi}{20\pi} = \dfrac{1}{10} \text{ m} = 10 \text{ cm}$ Thus, the minimum distance between two points having the same speed is: $\text{Distance} = \dfrac{\lambda}{2} = \dfrac{10 \text{ cm}}{2} = 5 \text{ cm}$

Q73

A heavy ball of mass M is suspendeed from the ceiling of a car by a light string of mass m (m < < M). When the car is at rest, the speed of transverse waves in the string is 60 ms

-

1. When the car has acceleration a, the wave-speed increases to 60.5 ms

-

1. The value of a, in terms of gravitational acceleration g, is closest to :

A

{g \over {30}}

B

{g \over 5}

C

{g \over 10}

D

{g \over 20}

Correct Answer

Option B

Solution

Resultant force on the ball of mass M when car is moving with a acceleration a is , Fnet =

\sqrt {{{\left( {Mg} \right)}^2} + {{\left( {Ma} \right)}^2}}

=

M\sqrt {{g^2} + {a^2}}

$\therefore$ T = M

\sqrt {{g^2} + {a^2}}

We know, Velocity, V =

\sqrt {{T \over \mu }}

When Car is at rest then, 60 =

\sqrt {{{Mg} \over \mu }}

. . . . (1) and when is moving then 60.5 =

\sqrt {{{M\sqrt {{g^2} + {a^2}} } \over \mu }}

. . . . (2) By dividing (2) by (1) we get,

{{60.5} \over {60}} = \sqrt {{{\sqrt {{g^2} + {a^2}} } \over g}}

$\Rightarrow$

\left( {1 + {{0.5} \over {60}}} \right)

=

{\left( {{{{g^2} + {a^2}} \over {{g^2}}}} \right)^{{1 \over 4}}}

$\Rightarrow$

{{{g^2} + {a^2}} \over {{g^2}}}

=

{\left( {1 + {{0.5} \over {60}}} \right)^4}

$\Rightarrow$

{{{g^2} + {a^2}} \over {{g^2}}}

= 1 + 4 $\times$

{{{0.5} \over {60}}}

[Using Binomial approximation] $\Rightarrow$

{{{g^2} + {a^2}} \over {{g^2}}}

= 1 +

{1 \over {30}}

$\Rightarrow$ 1 +

{{{a^2}} \over {{g^2}}}

= 1 +

{1 \over {30}}

$\Rightarrow$

{a \over g}

=

{1 \over {\sqrt {30} }}

$\Rightarrow$ a =

{g \over {\sqrt {30} }}

$\therefore$ Closest answer, a =

{g \over 5}

Q74

A tuning fork A of unknown frequency produces 5 beats/s with a fork of known frequency 340 Hz. When fork A is filed, the beat frequency decreases to 2 beats/s. What is the frequency of fork A?

A 335 Hz

B 345 Hz

C 338 Hz

D 342 Hz

Correct Answer

Option A

Solution

Initially beat frequency = 5Hz so, $\rho$ A = 340 $\pm$ 5 = 345 Hz, or 335 Hz after filing frequency increases slightly so, new value of frequency of A > $\rho$ A Now, beat frequency = 2Hz $\Rightarrow$ new $\rho$ A = 340 $\pm$ 2 = 342 Hz, or 338 Hz hence, original frequency of A is $\rho$ A = 335 Hz

Q75

A uniform string of length

20

m

is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the supports is : (take

{\,\,g = 10m{s^{ - 2}}}

)

A

2\sqrt 2 s

B

2\pi \sqrt 2 s

C

2\pi \sqrt 2 s

D

2

s

Correct Answer

Option A

Solution

We know that velocity in string is given by

v = \sqrt {{T \over \mu }} \,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)

where

\mu = {m \over {\rm I}} = {{mass\,\,\,of\,\,\,string} \over {length\,\,\,of\,\,\,string}}

The tension

T = {m \over \ell } \times x \times g\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

From

(a)

and

(b)

{{dx} \over {dt}} = \sqrt {gx}

{x^{ - 1/2}}\,dx = \sqrt g \,dt

$\therefore$

\int\limits_0^\ell {{x^{ - 1/2}}} dx - \sqrt g \int\limits_0^\ell {dt}

2\sqrt \ell = \sqrt g \times t

$\therefore$

t = 2\sqrt {{\ell \over g}} = 2\sqrt {{{20} \over {10}}} = 2\sqrt 2

Q76

A wave

y=a

\sin \left( {\omega t - kx} \right)

on a string meets with another wave producing a node at

x=0.

Then the equation of the unknown wave is

A

y = a\,\sin \,\left( {\omega t + kx} \right)

B

y = - a\,\sin \,\left( {\omega t + kx} \right)

C

y = a\,\sin \,\left( {\omega t - kx} \right)

D

y = - a\,\sin \,\left( {\omega t - kx} \right)

Correct Answer

Option B

Solution

To form a node there should be superposition of this wave with the reflected wave.

The reflected wave should travel in opposite direction with a phase change of $\pi$ .

The equation of the reflected wave will be

y = a\sin \left( {\omega t + kx + \pi } \right)

\Rightarrow y = - a\sin \left( {\omega t + kx} \right)

Q77

A uniform thin rope of length 12 m and mass 6 kg hangs vertically from a rigid support and a block of mass 2 kg is attached to its free end. A transverse short wavetrain of wavelength 6 cm is produced at the lower end of the rope. What is the wavelength of the wavetrain (in cm) when it reaches the top of the rope ?

A 12

B 3

C 9

D 6

Correct Answer

Option A

Solution

T1 = 2g T2 = 8g V =

\sqrt {{T \over \mu }}

$\therefore$ V $\propto$

\sqrt T

Also V = f $\lambda$ $\therefore$ V1 = f1 $\lambda$ 1 and V2 = f2 $\lambda$ 2 We know frequency of sources are same. $\therefore$ f1 = f2 So

\sqrt {{{{T_1}} \over {{T_2}}}} = {{{\lambda _1}} \over {{\lambda _2}}}

$\Rightarrow$

\sqrt {{{2g} \over {8g}}} = {6 \over {{\lambda _2}}}

$\Rightarrow$ $\lambda$ 2 = 12 cm

Q78

A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound ? [Given reference intensity of sound as 10–12 W/m2 ]

A 20 cm

B 10 cm

C 40 cm

D 30 cm

Correct Answer

Option C

Solution

Sound level = 10

{\log _{10}}\left( {{I \over {{I_0}}}} \right)

$\Rightarrow$ 120 = 10

{\log _{10}}\left( {{I \over {{{10}^{ - 12}}}}} \right)

$\Rightarrow$ 12 =

{\log _{10}}\left( {{I \over {{{10}^{ - 12}}}}} \right)

$\Rightarrow$ 1012 =

{{I \over {{{10}^{ - 12}}}}}

$\Rightarrow$ I = 1 W/m2 Also we know, I =

{P \over {4\pi {r^2}}}

$\Rightarrow$ 1 =

{2 \over {4\pi {r^2}}}

$\Rightarrow$ r = 40 cm

Q79

When temperature increases, the frequency of a tuning fork

A increases

B decreases

C remains same

D increases or decreases depending on the material

Correct Answer

Option B

Solution

KEY CONCEPT : The frequency of a tuning fork is given by the expression

f = {{{m^2}k} \over {4\sqrt 3 \pi {\ell ^2}}}\sqrt {{Y \over \rho }}

As temperature increases,

\ell

increases and therefore

f

decreases.

Q80

A cylindrical tube, open at both ends, has a fundamental frequency,

f,

in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now :

A

f

B

f/2

C

3/4

D

2f

Correct Answer

Option A

Solution

The fundamental frequency of open tube

{v_0} = {v \over {2{l_0}}}\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

That of closed pipe

{v_c} = {\upsilon \over {4{l_c}}}\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

According to the problem

{l_c} = {{{l_0}} \over 2}

Thus

{v_c} = {\upsilon \over {{l_0}/2}} \Rightarrow {v_c}{\upsilon \over {2l}}\,\,\,\,...\left( {iii} \right)

From equations

(i)

and

(iii)

{v_0} = {v_c}

Thus,

{v_c} = f

\,\,\,\left( {\,\,} \right.

as

{v_0} = f

is given

\left. {\,\,} \right)