WAVE MOTION QUIZ-9

Q1. The wavelength of two notes in air are 36/195 m and 36/193m. each note produces 10 beats per second separately with a third note of fixed frequency. The velocity of sound in air in m/s is

•  330
•  340
•  350
•  360

Solution
Beat frequency=v_1~v_2 Let the frequency of third note be n. Then, 195v/36-v=10 …(i) And v-193v/36=10 …(ii) Adding Eqs. (i) and (ii) v/18=20 ⟹ v=360 ms^(-1)

Q2.Law of superposition is applicable to only

•  Light waves
•  All kinds of waves
•  Transverse waves
•  Sound waves

Solution
B

Q3.  A car moving with a velocity of 36 km^(-1) crosses a siren of frequency 500 Hz. The apparent frequency of siren after passing it will be

•   520 Hz
•  460 Hz
•  540 Hz
•  485 Hz

Solution
Given that, velocity of the car =36000m/3600s=10ms^(-1) v_emitted=500 Hz v_sound=330 ms^(-1) We know that =v_emitted×((v_sound-v_obs ))/c_sound Hence, observed frequency v_obs=500×(330-10)/330 v_obs=485Hz

Q4. Who wrote the famous Book “Constitution of India - Defaced and Defied”?

•  3.2 cm
•  3.2 m
•  4.2 cm
•  4.2 m

Solution
Minimum audible frequency =20 Hz ⇒v/4l=20⇒l=336/(4×20)=4.2 m

Q5.A whistle revolves in a circle with an angular speed of 20 rad/sec using a string of length 50 cm. If the frequency of sound from the whistle is 385 Hz, then what is the minimum frequency heard by an observer, which is far away from the centre in the same plane? (v=340 m/s)

•  374 Hz
•  385 Hz
•  394 Hz
•  333 Hz

Solution
 Minimum frequency will be heard, when whistle moves away from the listener. n_min=n(v/(v+v_s )) where v=rω=0.5×20=10 m/s ⇒n_min=385(340/(340+10))=374Hz

Q6. The velocity of sound I air is 330ms^(-1). The rms velocity of air molecules (γ=1.4) is approximately equal to

•  400 ms^(-1)
•  471.4 ms^(-1)
• 231 ms^(-1)
•  462 ms^(-1)

Solution
C_(rms=v) √(3/γ)=330×√(3/1.4)=471.4 ms^(-1)

Q7.For the stationary wave y=4 sin⁡〖(πx/15) cos⁡〖(96πt)〗 〗, the distance between a node and the next antinode is

•  22.5
•  15
•  7.5
•  30

Solution
Comparing given equation with standard equation y=2a sin⁡〖2πx/λ〗 cos⁡〖2πvt/λ〗 gives us 2π/λ=π/15⇒λ=30 Distance between nearest node and antinodes =λ/4=30/4=7.5

Q8.A source is moving towards an observer with a speed of 20 m/s and having frequency of 240 Hz. The observer is now moving towards the source with a speed of 20 m/s. Apparent frequency herad by observer, if velocity of sound is 340 m/s, is

•  240 Hz
•  270 Hz
•  280 Hz
•  360 Hz

Solution
n^'=n((v+v_O)/(v-v_S ))=240((340+20)/(340-20))=270 Hz

Q9.A tuning fork of frequency 340 Hz is vibrated just above the tube of 120 cm height. Water is poured slowly in the tube, what is the minimum height of water necessary for the resonance?

•  30 cm
•  25 cm
•  35 cm
•  45 cm

Solution
Using relation v=vλ Or λ=v/v=340/340-1m If length of resonance columns are l_1,l_2 and l_3, then l_1=λ/4=1/4 m=25 cm (for first resonance) l_2=3λ/4=75cm (for second resonance) l_3=5λ/4=125 cm for third resonance This case of third resonance is impossible because total length of the tube is 120 cm So, minimum height of water =120-75=45cm

Q10. Two waves are represented by y_1=4 sin⁡404πt and y_2=3 sin⁡400πt. Then

•  Beat frequency is 4 Hz and the ratio of maximum to minimum intensity is 49∶1
•  Beat frequency is 2 Hz and the ratio of maximum to minimum intensity is 49∶1
•  Beat frequency is 2 Hz and the ratio of maximum to minimum intensity is 1∶49
• Beat frequency is 4 Hz and the ratio of maximum to minimum intensity is 1∶49

Solution
Given ∶y_1=4 sin⁡〖404πt,y_2=3 sin⁡400πt 〗 ∴ω_1=404π,ω_2=400π,A_1=4,A_2=3 ω_1=2πv_1⇒404π=2πv_1⇒v_1=202 Hz ω_2=2πv_2⇒400π=2πv_2⇒v_2=200 Hz Beat frequency =v_1-v_2=202-200=2 Hz I_max/I_min =((A_1+A_2)/(A_1-A_2 ))^2=((4+3)/(4-3))^2=(7/1)^2=49/1