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A string of length 0.4 m and mass 10^(-2...

A string of length `0.4 m` and mass `10^(-2)kg` is tightly clamped at its ends. The tension in the string is `1.6 N`. Identical wave pulse are produced at one end at equal intervals of time, `Deltat`. The minimum value of `Deltat` which allows constructive interference of successive pulse is

A

`0.05` s

B

`0.40` s

C

`0.20` s

D

`0.10` s

Text Solution

Verified by Experts

The correct Answer is:
d

The speed of transverse wave in a flexible stretched
string depends upon the tension T in the string and the
mass per unit length `(mu)` as
`v=sqrt(T/mu)=sqrt(T/(m//l)`
Putting the numerical values form the question,
`therefore v = sqrt(1.6/(10^(-2)//0.4))=8" ms"^(-1)`
We know that when pulse is reflected from the other end
it suffers a phase change of `180^(@)`. Again when the
reflection takes place from first end, a phase change of
`180^(@)` occurs. Therefore, after successive reflections same
wavve pulse is obtained as previous. If at this instant
another same wave pulse is obtained, then these two
wave pulses produce constructive interference.
Hence, for constructive interference between successive
pulses `Delta t_(min)= (0.4+0.4)/8 =0.10` s
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-STATIONARY WAVES -EXERCISE 1
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  18. Two stratched strings of same material are vibrating under some tensio...

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  19. A uniform wire of length L, diameter D and density rho is stretched un...

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