The driver of a bus approaching a big wall notices that the frequency of his bus’s horn changes from 420 Hz to 490 Hz ….

Q: The driver of a bus approaching a big wall notices that the frequency of his bus’s horn changes from 420 Hz to 490 Hz when he hears it after it gets reflected from the wall . Find the speed of bus if speed of the sound is 330 m/s

(a) 91 km/h

(b) 61 km/h

(c) 81 km/h

(d) 71 km/h

Click to See Solution :
Ans: (a)
Sol: Frequency appeared at wall

$\displaystyle f’ = \frac{v}{v-v_b}\times f $

$\displaystyle f’ = \frac{330}{330-v_b}\times f $ …(i)

Now Frequency of reflected sound ,

$\displaystyle f” = \frac{330 + v_b}{330}\times f’ $

$\displaystyle f” = \frac{330 + v_b}{330}\times \frac{330}{330-v_b}\times f $ from(i)

$\displaystyle f” = \frac{330 + v_b}{330-v_b}\times f $

$\displaystyle 490 = \frac{330 + v_b}{330-v_b}\times 420 $

vb = 25.4 m/s = 91 km/h

 

In a resonance tube experiment the tube is filled with water upto a height of 17.0cm from bottom , it resonates with ….

Q: In a resonance tube experiment the tube is filled with water upto a height of 17.0cm from bottom , it resonates with a given tuning fork . When the water level is raised the next resonance with same tuning fork occurs at a height of 24.5 cm . If velocity of sound in air is 330 m/s , the tuning fork frequency is

(a) 2200 Hz

(b) 550 Hz

(c) 1100 Hz

(d) 3300 Hz

Click to See Solution :
Ans: (a)

Sol: A resonance tube is an organ pipe closed at one end . If L1 and L2 be the first and second resonance lengths with a tuning fork of frequency ν , then the speed of sound in air is given by

$\large v = 2 \nu (L_2 – L_1)$

$\large \nu \lambda = 2 \nu (L_2 – L_1)$

$\large \lambda = 2 (L_2 – L_1)$

$\large \lambda = 2 (24.5-17 – L_1)$

$\large \lambda = 15 cm $

$\large v = \nu \lambda $

$\large 300 = \nu \times 15 \times 10^{-2}$

ν = 2200 Hz

 

For a transverse wave travelling along a straight line , the distance between two peaks (crests) is 5 m ….

Q: For a transverse wave travelling along a straight line , the distance between two peaks (crests) is 5 m , while the distance between one crest and one trough is 1.5 m . The possible wavelength (in m) of the waves are

(a) 1/1 , 1/3 , 1/5 , …..

(b) 1 ,2 ,3 ,…..

(c) 1,3 , 5 ….

(d) 1/2 , 1/4 , 1/6 ,…

Click to See Solution :
Ans: (a)
Sol: Given the distance between two crests is 5 m and between any crest and trough is 1.5 m . From given options , which satisfies these two conditions is option(a)

 

A transverse wave travels on a taut steel wire with a velocity of v when tension in it is 2.06 × 10^4 N …..

Q: A transverse wave travels on a taut steel wire with a velocity of v when tension in it is 2.06 × 104 N . When the tension is changed to T , the velocity changed to v/2 . The value of T is close to

(a) 10.2 × 102 N

(b) 5.15 × 103 N

(c) 2.50 × 104 N

(d) 30.5 × 104 N

Click to See Solution :
Ans: (b)
Sol: v1 = v , T1 = 2.06 × 104 N

v2 = v/2 , T2 = ?

$\large v \propto \sqrt{T}$

$\displaystyle \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$

$\displaystyle \frac{v}{2 v} = \sqrt{\frac{T_2}{T_1}}$

$\displaystyle \frac{1}{4} = \frac{T_2}{T_1} $

$\displaystyle T_2 = \frac{T_1}{4} = \frac{2.06 \times 10^4}{4}$

= 5.15 × 103 N

 

A stationary observer receives sound from two identical tuning forks , one of which approaches and the other one recedes ….

Q: A stationary observer receives sound from two identical tuning forks , one of which approaches and the other one recedes with the same speed (much less than speed of sound speed) . The observer hears 2 beats/sec . The oscillation frequency of each tuning fork is νo = 1400 Hz and the velocity of sound in air is 350 m/s . The speed of each tuning fork is close to

(a) 1/8 m/s

(b) 1/4 m/s

(c) 1 m/s

(d) 1/2 m/s

Click to See Solution :
Ans: (b)

Sol: Let v = Speed of tuning fork

$\displaystyle \nu_1 = \frac{v_s}{v_s – v} \nu_o $

$\displaystyle \nu_2 = \frac{v_s}{v_s + v} \nu_o $

Beat Frequency = ν1 – ν2

$\displaystyle 2 = v_s \times \nu_o [ \frac{1}{v_s – v} – \frac{1}{v_s + v}] $

$\displaystyle 2 = v_s \times \nu_o [ \frac{2 v}{v_s^2 – v^2}] $

v << vs , so neglecting v2

$\displaystyle 2 = \frac{ 1400 \times 350 \times 2 v}{350 \times 350} $

v = 1/4 m/s