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Standing Waves On a String LAB:

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Answer #1

1] Increasing frequency of the will effect by

v= f* lamda

= v = remains constant

frequency increases simutaneously lamda decreases so

no. of harmonics shall increase as we increase the frequency

2] Tension is directly proportional to the velocity

v = sqrt[T/mue]

so on decreasing tension basically means decreasing velocity

v= f* lamda

= decreasing v

means we are decreasing the harmonics

3] from part 2

increasing T means increasing velocity

which results in increasing frequency

increasing the harmonics

4]increasing the frequency will not effect on the velocity

5] Force = newton = Kgms^-2

so units for question 5 are

L = m

mue = Kgm^-1

frequency = s^-1

n= unitless quantity

so placing all the values in the equation given we get

= m^2 * s^-2 * Kg* m^-1

= Kgms^-2

which is the unit of the force

6] In the equation given in 6

we see

frequency has unit = s^-1

so the equation given should have this unit too

so

= SQRT[Kgms^-2/m^2* Kgm^-1 ]

i placed the values of the unit

if doubt pls ask

=which comes out to be

= SQRT[s^-2]

= s^-1

ANY DOUBT PLS ASK IT WS GREAT TO HELP YOU !! PLS RATE THE TWO QUESTIONS !

HAPPY TO HELP YOU !

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Answer #3

1) frequency= nv/2l where n=1,2,3 v=velocity of sound l=length of string

for a standing wave frequency of n harmonic=nx frequency of 1st harmonics

so increasing frequency will increase the harmonics

2)frequency directly proportional to velocity of sound which is directly proportional to square root of tension

so decrasing tension will decrease the frequency and hence the harmonics.

3)increasing T will increase frequency.

4)increasing f will have no effect on v

5)4xml2f2/nl=mlf2=[MLT-2]= dimensions of force

6)(Tl/4l2m)1/2=(MLT-2L/L2M)1/2=[T-2]1/2=T-1=frequency

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