Waves MCQ Questions & Answers in Oscillation and Mechanical Waves | Physics
Learn Waves MCQ questions & answers in Oscillation and Mechanical Waves are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
201.
Two pulses in a stretched string whose centers are initially $$8\,cm$$ apart are moving towards each other as shown in the figure. The speed of each pulse is $$2\,cm/s.$$ After 2 seconds, the total energy of the pulses will be
A
zero
B
purely kinetic
C
purely potential
D
partly kinetic and partly potential
Answer :
purely kinetic
After two seconds pulses will overlap each other. NOTE : According to superposition principle the string will not have any distortion and will be straight.
Hence there will be no P.E. The total energy will be only kinetic.
202.
Two waves are represented by the equations $${y_1} = a\sin \left( {\omega t + kx + 0.57} \right)m$$ and $${y_2} = a\cos \left( {\omega t + kx} \right)m,$$ where $$x$$ is in metre and $$t$$ in second. The phase difference between them is
203.
A man is watching two trains, one leaving and the other coming in with equal speeds of $$4\,m/\sec.$$ If they sound their whistles, each of frequency $$240\,Hz,$$ the number of beats heard by the man (velocity of sound in air $$= 320\,m/\sec$$ ) will be equal to
A
6
B
3
C
0
D
12
Answer :
6
Frequency of Sound heard by the man from approaching train.
$${n_a} = n\left( {\frac{v}{{v - {v_s}}}} \right) = 240\left( {\frac{{320}}{{320 - 4}}} \right) = 243\,Hz$$
Frequency of sound heard by the man from receding train $${n_r} = n\left( {\frac{v}{{v + {v_s}}}} \right) = 240\left( {\frac{{320}}{{320 + 4}}} \right) = 237\,Hz$$
Hence, number of beats heard by man per $$\sec$$
$$ = {n_a} - {n_r} = 243 - 237 = 6$$
204.
Two waves of same frequency and intensity superimpose on each other in opposite phases. After the superposition, the intensity and frequency of waves will
A
increase
B
decrease
C
remain constant
D
become zero
Answer :
become zero
Interference phenomenon is common to sound and light. In sound, the interference is said to be constructive at points where resultant intensity is maximum (are in phase) and destructive at points where resultant intensity is minimum or zero (are in opposite phase).
205.
When a sound wave goes from one medium to another, the quantity that remains unchanged is
A
frequency
B
amplitude
C
wavelength
D
speed
Answer :
frequency
Frequency of wave is a function of the source of waves. Therefore, it remains unchanged.
206.
Two waves are said to be coherent, if they have
A
same phase but different amplitude
B
same frequency but different amplitude
C
same frequency, phase and amplitude
D
different frequency, phase and amplitude
Answer :
same frequency, phase and amplitude
Two wave are said to be coherent, when they have same frequency, amplitude and constant phase difference.
207.
A string is stretched between fixed points separated by $$75.0\,cm.$$ It is observed to have resonant frequencies of $$420\,Hz$$ and $$315\,Hz.$$ There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is
A
$$105\,Hz$$
B
$$1.05\,Hz$$
C
$$1050\,Hz$$
D
$$10.5\,Hz$$
Answer :
$$105\,Hz$$
$$\eqalign{
& {\text{Given}}\,\,\frac{{nv}}{{2\ell }} = 315\,\,{\text{and}}\,\,\left( {n + 1} \right)\frac{v}{{2\ell }} = 420 \cr
& \Rightarrow \frac{{n + 1}}{n} = \frac{{420}}{{315}} \Rightarrow n = 3 \cr
& {\text{Hence}}\,\,3 \times \frac{v}{{2\ell }} = 315 \Rightarrow \frac{v}{{2\ell }} = 105\,Hz \cr} $$
The lowest resonant frequency is when $$n = 1$$
Therefore lowest resonant frequency $$= 105\,Hz.$$