A string of length L and mass M hangs freely from a fixed point. Then, which of the following are correct.a) the time taken by a transverse pulse to traverse the string is (Lg)b) the time taken by a transverse pulse to traverse the string is 2(Lg)c) the velocity of the transverse wave along the string at a point at a distance x from the lower end is (2gx).d) the velocity of the transverse wave along the string at a point at a distance x from the lower end is (gx).

Total length of string =L Mass of the string =M The velocity of transverse wavev=Tm where, T= weight of part of string hanging below the point under consideration =(ML)xg ⇒v=(ML)xg(ML)=xg

A string of length L and mass M hangs freely from a fixed point. Then, which of the following are correct.a) the time taken by a transverse pulse to traverse the string is (Lg)b) the time taken by a transverse pulse to traverse the string is 2(Lg)c) the velocity of the transverse wave along the string at a point at a distance x from the lower end is (2gx).d) the velocity of the transverse wave along the string at a point at a distance x from the lower end is (gx).

Total length of string =L Mass of the string =M The velocity of transverse wavev=Tm where, T= weight of part of string hanging below the point under consideration =(ML)xg ⇒v=(ML)xg(ML)=xg

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A string of length L and mass M hangs freely from a fixed point. Then, which of the following are correct.
a) the time taken by a transverse pulse to traverse the string is $(\frac{L}{g})$
b) the time taken by a transverse pulse to traverse the string is $2 \sqrt{(\frac{L}{g})}$
c) the velocity of the transverse wave along the string at a point at a distance x from the lower end is $\sqrt{(2 gx)}$ .
d) the velocity of the transverse wave along the string at a point at a distance x from the lower end is $\sqrt{(gx)}$ .

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a and b

a and c

c and d

b and d

answer is D.

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Detailed Solution

$T o t a l l e n g t h o f s t r i n g = L M a s s o f t h e s t r i n g = M The velocity of transverse wave v = \sqrt{\frac{T}{m}} w h e r e, T = w e i g h t o f p a r t o f s t r i n g h a n g i n g b e l o w t h e p o i n t u n d e r c o n s i d e r a t i o n = (\frac{M}{L}) xg \Rightarrow v = \sqrt{\frac{(\frac{M}{L}) xg}{(\frac{M}{L})}} = \sqrt{xg}$