# 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

**Asked By: LETTERTOKNOW**6 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**6 year ago

**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**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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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 H_{2}O 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**7 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 s_{1 }&s_{2} reach at a point P in phase. resultant loudness at a point P is ndB higher than loudness of s_{1} the value of n =?

ANS) 6

**Asked By: SARIKA**7 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/I_{0} due to a single source

and loudness at the point where 2 waves meet in phase = 10log4I/I_{0 }= 10logI/I_{0}+ 10log4

since 10log4 = 6 so increment = 6

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*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**7 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**7 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