6 - Wave in a string Questions Answers

A small ring is threaded on an inextensible frictionless cord of length 2l.  The ends of the cord are fixed to a horizontal ceiling.  In equilibrium, the ring is at a depth h below the ceiling.  Now the ring is pulled aside by a small distance in the vertical plane containing the cord and released.  If period of small oscillations of the ring is kπ, find k.  Given, 


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Joshi sir comment

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



3) 54>X>36

4) 36>X>18

ANS (2)

AIEEE 2008

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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  


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




4) f

aieee 2012

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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
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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?



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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.

a sonometer wire supports a 4kg load and vibrates in fundamental mode with a tuning fork of frequency 416 Hz. the length of the wire between the bridges is now doubled. in order to maintain fundamental mode the load should be changed to

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Joshi sir comment

for sonometer wire 

in fundamental mode l = λ/2 = v/2f       here l, λ and f are length , wavelength and frequency respectively

now v = √(T/µ)      here T and µ are tension and mass per unit length

so finally we will get l = √(T/µ)/2f

make two equations by using this formula and given conditions

in the two equations f and µ will be same

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