A string of length L and mass M hangs freely from a fixed point. Then, the velocity of transverse waves along the string at a distance x from the free end is

Question

A string of length L and mass M hangs freely from a fixed point. Then, the velocity of transverse waves along the string at a distance x from the free end is (A)  gx  (B)  √gx  (C)  gL  (D)  √gL

Tardigrade Admin · Accepted Answer

Total length of string = L Mass of the string = M
The velocity of transverse wave

v = \frac{T}{m}

where,

T =

weight of part of string hanging below the point under consideration

= (\frac{M}{L 1}) xg

\Rightarrow

v = \frac{( \frac{M}{L} ) xg}{( \frac{M}{L} )} = xg

Q. A string of length L and mass M hangs freely from a fixed point. Then, the velocity of transverse waves along the string at a distance $x$ from the free end is

$gx$

$gx$

$gL$

$gL$

Total length of string = L Mass of the string = M
The velocity of transverse wave
$v = \frac{T}{m}$
where, $T =$ weight of part of string hanging below the point under consideration
$= (\frac{M}{L 1}) xg$
$\Rightarrow$ $v = \frac{( \frac{M}{L} ) xg}{( \frac{M}{L} )} = xg$

Q. A string of length L and mass M hangs freely from a fixed point. Then, the velocity of transverse waves along the string at a distance x from the free end is

gx

gx​

gL

gL​

Total length of string = L Mass of the string = M The velocity of transverse wave v=mT​​ where, T= weight of part of string hanging below the point under consideration =(L1M​)xg ⇒ v=(LM​)(LM​)xg​​=xg​

Q. A string of length L and mass M hangs freely from a fixed point. Then, the velocity of transverse waves along the string at a distance $x$ from the free end is

$gx$

$gx$

$gL$

$gL$

Total length of string = L Mass of the string = M
The velocity of transverse wave
$v = \frac{T}{m}$
where, $T =$ weight of part of string hanging below the point under consideration
$= (\frac{M}{L 1}) xg$
$\Rightarrow$ $v = \frac{( \frac{M}{L} ) xg}{( \frac{M}{L} )} = xg$