Waves — Page 4 | JEE Physics

Q31

In an experiment to determine the period of a simple pendulum of length 1 m, it is attached to different spherical bobs of radii r1 and r2 . The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be 5×10−4 s, the difference in radii,

\left| {} \right.

r1

-

r2

\left| {} \right.

is best given by :

A 1 cm

B 0.05 cm

C 0.5 cm

D 0.01 cm

Correct Answer

Option B

Solution

The time period is given by the formula

T = 2\pi \sqrt {{l \over g}}

which clearly indicates that the time period is directly proportional to the length of the pendulum l, that is,

T \propto \sqrt l

Here, l = 1 m. Therefore,

{{\Delta T} \over l} = {1 \over 2}{{\Delta l} \over l}

....... (1) where

\Delta l = \left| {{r_1} - {r_2}} \right|

. Substituting the values in Eq. (1), we get

5 \times {10^{ - 4}} = {1 \over 2}\left( {{{{r_1} - {r_2}} \over 1}} \right)

\Rightarrow {r_1} - {r_2} = 10 \times {10^{ - 4}} = {10^{ - 3}}

m

= {10^{ - 1}}

cm

= 0.1

cm

Q32

The end correction of a resonance column is 1 cm. If the shortest length resonating with the tunning fork is 10 cm, the next resonating length should be :

A 28 cm

B 32 cm

C 36 cm

D 40 c

Correct Answer

Option B

Solution

Given, End correction (e) = 1 cm For first resonance,

{\lambda \over 4} = {l_1} + e

= 10 + 1 = 11 cm For second resonance,

{3\lambda \over 4} = {l_2} + e

$\Rightarrow$

{l_2}

= 3 $\times$ 11 - 1 = 32 cm

Q33

5 beats / econd are heard when a tuning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either 0.95 m or 1 m. The frequency of the fork will be :

A 195 Hz

B 150 Hz

C 300 Hz

D 251 Hz

Correct Answer

Option A

Solution

Length of wire is L1 = 0.95 m; L2 = 1 m. Number of beats per second heard = 5 Let frequency of fork be f. Therefore,

{v \over {2{L_1}}} - f = 5 \Rightarrow {v \over {2{L_1}}} = 5 + f

...... (1) and

f - {v \over {2{L_2}}} = 5 \Rightarrow {v \over {2{L_2}}} = f - 5

.... (2) Dividing Eq. (1) by Eq. (2), we get

{{{v \over {2{L_1}}}} \over {{v \over {2{L_2}}}}} = {{5 + f} \over {f - 5}} \Rightarrow {{{L_2}} \over {{L_1}}} = {{f + 5} \over {f - 5}}

\Rightarrow {1 \over {0.95}} = {{f + 5} \over {f - 5}} \Rightarrow f - 5 = 0.95f + 4.75

\Rightarrow f - 0.95f = 5 + 4.75 \Rightarrow 0.05f = 9.75 \Rightarrow f = {{9.75} \over {0.05}}

\Rightarrow f = 195

Hz

Q34

Two sources of sound S1 and S2 produce sound waves of same frequency 660 Hz. A listener is moving from source S1 towards S2 with a constant speed u m/s and he hears 10 beats/s. The velocity of sound is 330 m/s. Then, u equals :

A 10.0 m/s

B 5.5 m/s

C 15.0 m/s

D 2.5 m/s

Correct Answer

Option D

Solution

As observer goes away from source S1 so apparent frequency,

{f_1} = \left( {{{v - u} \over v}} \right)f

here

v

= speed of sound,

u

= speed of observer As observer goes towards source S2 so apparent frequency,

{f_2} = \left( {{{v + u} \over v}} \right)f

Beat frequency =

{f_2} - {f_1}

= 10 $\Rightarrow$

\left( {{{v + u} \over v}} \right)f

-

\left( {{{v - u} \over v}} \right)f

= 10 $\Rightarrow$

f\left( {{{v - u - v + u} \over v}} \right)

= 10 $\Rightarrow$

f\left( {{{2u} \over v}} \right)

= 10 $\Rightarrow$

u = {{10v} \over {2f}}

=

{{10 \times 330} \over {2 \times 660}}

= 2.5 m/s

Q35

A submarine (A) travelling at 18 km/hr is being chased along the line of its velocity by another submarine (B) travelling at 27 km/hr. B sends a sonar signal of 500 Hz to detect A and receives a reflected sound of frequency

\upsilon

. The value of

\upsilon

is close to: (Speed of sound in water =1500 ms–1)

A 507 Hz

B 502 Hz

C 499 Hz

D 504 Hz

Correct Answer

Option B

Solution

f1 (frequency received by A)

= {v_0}\left[ {{{1500 - 5} \over {1500 - 7.5}}} \right]

f2 [frequency received by B]

= {v_0}{{1495} \over {1492.5}} \times {{1507.5} \over {1505}}

= 502\,Hz

Q36

A stationary source emits sound waves of frequency 500 Hz. Two observers moving along a line passing through the source detect sound to be of frequencies 480 Hz and 530 Hz. Their respective speeds are, in ms–1, (Given speed of sound = 300 m/s)

A 12, 18

B 16, 14

C 12, 16

D 8, 18

Correct Answer

Option A

Solution

v = {{v + {v_o}} \over v}{v_o}

\Rightarrow {v_o} = \left( {{v \over {{v_o}}} - 1} \right)v

{v_o} = \left( {{{530} \over {500}} - 1} \right)300 = 18\,m/s

{v_o} = \left| {\left( {{{480} \over {500}} - 1} \right)300} \right| = 12\,m/s

Q37

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

Q38

A musician using an open flute of length 50 cm producess second harmonic sound waves. A person runs towards the musician from another end of hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to :

A 666 Hz

B 753 Hz

C 500 Hz

D 333 Hz

Correct Answer

Option A

Solution

Frequency of sound wave produce by flute =

{{2{V_S}} \over {2\ell }}

=

{{2 \times 330} \over {2 \times 50 \times {{10}^{ - 2}}}}

= 660 Hz Speed of the observer = 10km/hr = 10 $\times$

{5 \over {18}}

m/s =

{{25} \over 9}

m/s Frequency heard by the observer, f' =

\left( {{{{V_S} + {V_o}} \over {{V_S}}}} \right)f

=

\left( {{{330 + {{25} \over 9}} \over {330}}} \right) \times 660

= 666 Hz

Q39

A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is ƒ1. If the speed of the train is reduced to 17 m/s, the frequency registered is ƒ2. If speed of sound is 340 m/s, then the ratio ƒ1/ƒ2 is -

A 19/18

B 20/19

C 21/20

D 18/17

Correct Answer

Option A

Solution

fapp = f0

\left[ {{{{v_2} \pm {v_0}} \over {{v_2} \pm {v_s}}}} \right]

f1 = f0

\left[ {{{340} \over {340 - 34}}} \right]

f2 = f0

\left[ {{{340} \over {340 - 17}}} \right]

{{{f_1}} \over {{f_2}}} = {{340 - 17} \over {340 - 34}} = {{323} \over {306}} \Rightarrow {{{f_1}} \over {{f_2}}} = {{19} \over {18}}

Q40

A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to -

A 16.6 cm

B 10.0 cm

C 20.0 cm

D 33.3 cm

Correct Answer

Option C

Solution

Velocity of wave on string

V = \sqrt {{T \over \mu }} = \sqrt {{8 \over 5} \times 1000} = 40m/s

Now, wavelength of wave

\lambda = {v \over n} = {{40} \over {100}}m

Separation b/w successive nodes,

{\lambda \over 2} = {{20} \over {100}}\,m

$=$ 20 cm