10 - Wave in a string Questions Answers

Acceleration of point x=2m of a transverse wave is 5ĵ m/s^2 (5 m/s^2 in +ve y direction) and slope of waveform is given by dy/dx = 5sin[(pi/12)x] {where pi=22/7}. What is the speed of the particle at x=2?

Asked By: LETTERTOKNOW 5 year ago
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Acceleration of point x=2m of a transverse wave is 5ĵ m/s^2 (5 m/s^2 in +ve y direction) and slope of waveform is given by dy/dx = 5sin[(pi/12)x] {where pi=22/7}.

Asked By: LETTERTOKNOW 5 year ago
is this question helpfull: 4 0 submit your answer

Acceleration of point x=2m of a transverse wave is 5ĵ m/s^2 (5 m/s^2 in +ve y direction) and slope of waveform is given by dy/dx = 5sin[(pi/12)x] {where pi=22/7}.

Asked By: LETTERTOKNOW 5 year ago
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a wire of desity 900 kg/m3 is strechded between two clamps 1m apart while subjected to an extention of 0.05 cm.what is the lowest frequency of the transverse vibration in the wire.assuming young's modulus y=9×10 dyne/m2.

 

Asked By: DAWIT SOLOMON 6 year ago
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a wire of desity 900 kg/m3 is strechded between two clamps 1m apart while subjected to an extention of 0.05 cm.what is the lowest frequency of the transverse vibration in the wire.assuming young's modulus y=9×10 dyne/m2

 

Asked By: DAWIT SOLOMON 6 year ago
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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 6 year ago
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Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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

1)f/2

2)3f/4

 3)2f

4) f

aieee 2012

Asked By: SARIKA SHARMA 6 year ago
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Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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 6 year ago
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Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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 6 year ago
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Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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

Asked By: HIMANSHU MITTAL 6 year ago
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Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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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