22 - Waves and Oscillations Questions Answers

while measuring the speed of sound by performing a resonance column expt , a student gets d first resonance conditn at a column length of 18 cm during winter . Repeating d same expt during summer , she measure d column length to be x cm for the second resonance. THEN

1)18>X

2)X>54

3) 54>X>36

4) 36>X>18

ANS (2)

AIEEE 2008

Asked By: SARIKA SHARMA
is this question helpfull: 8 1 submit your answer
Joshi sir comment

since l α λ α v/n          n is frequency

on increasing temperature, v increases so λ increases

for first resonance l = λ/4

and for third resonance L = 3λ'/4

and λ> λ so x will be more than 54  

 

when a tuning fork vibrates with 1.0 m  or  1.05m long wire of a sonometer, 5 beats per second are produced in each case. what will be the frequency of the  tuning fork ??

Asked By: SARIKA SHARMA
is this question helpfull: 11 7 submit your answer
Joshi sir comment

for sonometer wire

l = λ/2

or 2l = v/n                                    here n represents frequency

or n = v/2l

for first wire of 1 m length n= v/2

for second wire of 1.05 m length n=v/2.1

let frequency of tuning fork = n1

so according to the given condition v/2 = n1+ 5

and v/2.1 = n1 - 5

solve now

 

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

Asked By: SARIKA SHARMA
is this question helpfull: 5 1 submit your answer
Joshi sir comment

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

Asked By: SARIKA SHARMA
is this question helpfull: 3 1 submit your answer
Joshi sir comment

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

Asked By: SARIKA
is this question helpfull: 4 0 submit your answer
Joshi sir comment

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/I= 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?

 

 

Asked By: SHUBHAM VED
is this question helpfull: 2 0 submit your answer
Joshi sir comment

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  ?

Asked By: SHUBHAM VED
is this question helpfull: 5 0 submit your answer
Joshi sir comment

 

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 . 

Asked By: AMIT DAS
is this question helpfull: 2 0 submit your answer
Joshi sir comment

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

Asked By: HIMANSHU MITTAL
is this question helpfull: 16 2 submit your answer
Joshi sir comment

I1/I= 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

Asked By: HIMANSHU MITTAL
is this question helpfull: 8 4 submit your answer
Joshi sir comment

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

Login Here

Username / Email :

Password :

Register | Forget Password