30 - Waves and Oscillations Questions Answers
TWO PENDULUM HAVE TIME PERIODS T &5T/4. THEY STARTS SHM AT THE SAME TIME FROM THE MEAN POSITION . WHAT WILL BE THE PHASE DIFFERENCE B/W THEM AFTER D BIGGER PENDULUM COMPLETED ONE OSCILLATn ??
ans 900
When bigger pendulum will complete its one oscillation, the smaller one will complete [1+(1/4)] oscillation. so phase difference = 2π*(T/4)/T = π/2
a cylinderical tube open at both ends has a fundamental frequency f in air the tube is dipped vertically in H2O so that half of it is in H2O the fundamental frequency of the air column is
1)f/2
2)3f/4
3)2f
4) f
aieee 2012
for first case l = λ/2 so l = v/2f
for second case l/2 = λ/4 so l/2 = v/4f
so f in both case are same.
QUES TWO IDENTICAL SOUNDS s1 &s2 reach at a point P in phase. resultant loudness at a point P is ndB higher than loudness of s1 the value of n =?
ANS) 6
Let intensity of one source = I
then intensity at a point where two waves meet in phase = I+I+2√I√Icos0 = 4I
now loudness = 10logI/I0 due to a single source
and loudness at the point where 2 waves meet in phase = 10log4I/I0 = 10logI/I0+ 10log4
since 10log4 = 6 so increment = 6
a uniformrope of lenth 10 m and mass 12 kg hangs verticaly from a rigid support . a block of mass 4 kg is attached to the free end of the rope . atrans verse wave of wavelenth 00.03 m is prodused at the lower end . wavelenth of pulse when it reaches upper end will be?
at top point tension in the string = 16g
and at bottom point T = 4g
and v = √(T/m) here m is mass per unit length which is same throughout the rope.
first calculate v at both the points
frequency will be same throughout
so use the formula v = nλ twice
once at bottom and once at top point
then by diving these 2 formulae we will get the result
try it.
2 comparable masses m1 and m2 r orbiting about their bodycenter o at distance r1 and r2 from their body center respectively with a common period T. the squar of time period ?
According to the diagram
Gm1m2/(r1+r2)2 = m1r1ω2 = m2r2ω2
compare first part with either second or third part of the given equation for getting ω2 then find T2
how can we calculate ratio of intensities of two sound waves having unequal amplitudes (one having double the amplitude of the other - both are sinusodial waves) and the graph is displacement vs time .
Intensity of a wave depends on two power of amplitude and two power of frequency. Check the graph carefully for getting idea about frequency and details of amplitude are given in the question.
the ratio of intensities between two coherent sound sources is 4:1. the difference of loudness in decibel between maximum and minimum intensities, when they interfere in space is
(1) 10 log2
(2) 20 log3
(3) 10 log3
(4) 20 log2
I1/I2 = 4/1 so a1/a2 = 2/1 so amax/ amin = 2+1/2-1 = 3/1 so Imax/ Imin= 9/1
now loudness L = 10 log(I/I0)
so difference of loudness = L1-L2 = 10 log (9I/I) = 10 log9 = 20 log3
two coherent light sources A and B with seperation 2λ are placed the x-axis symmetrically about the origin. they emit light of wavelength λ. obtain the position of maximas on a circle of large radius R lying in the xy plane and with the center at origin
for maxima path diffrence should be integral multiple of λ.
It is given that two coherent light sources A and B with seperation 2λ are placed on the x-axis symmetrically about the origin, and we have to get the points of maxima in the circle so at the point where circle cuts the X axis (path diffrence = 2λ) and where circle cuts the Y axis (path difference = 0) will be the point of maxima. Besides it one point will also be present in the circle in first quadrant where path difference will be λ so it will be a maxima also. it means total 8 points are there
two sound waves having intensities 4 unit and 9 unit coming from two coherent sources arrive at a point along the same line simultaneously in same phase. resultant intensity at the point will be
Use the formule I = I1 + I2 + 2 √I1 √I2 cosθ
here θ = 0
two sound sources of intensity I and 4I are used in an interference experiment. find the intensity at points where the waves from the two sources are superimposed with a phase difference of
(1) zero
(2) π/2
93)π
Use the formule I = I1 + I2 + 2 √I1 √I2 cosθ
here θ = 0