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# A uniform rope of length $12\,m$ and mass $6\,kg$ hangs vertically from a rigid support. A block of mass $2\,kg$ is attached to the free end of the rope. A transverse pulse of wavelength $0.06\,m$ is produced at the lower end of the rope. What is the wavelength of the pulse when it reaches the top of the rope?(A) $0.06\,m$(B) $0.03\,m$(C) $0.12\,m$(D) $0.09\,m$

Last updated date: 24th Jul 2024
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Hint The wavelength can be determined by using the velocity relation, by using the velocity relation the wavelength can be determined, the velocity relation can be determined by using the velocity of the string formula, then the wave length is determined.
Useful formula:
The velocity of the wave on the string is given by,
$v = \sqrt {\dfrac{T}{m}}$
Where, $v$ is the velocity of the wave of the string, $T$ is the tension of the string and $m$ is the mass of the string.

Given that,
The length of the rope is given as, $L = 12\,m$,
The mass which hangs vertically from a rigid support is, $m = 6\,kg$,
The mass which is attached to the free end of the rope is, $m = 2\,kg$,
The wavelength of the transverse pulse is given as, ${\lambda _1} = 0.06\,m$,
Now,
The velocity of the wave on the string is given by,
$v = \sqrt {\dfrac{T}{m}} \,.....................\left( 1 \right)$
The velocity relation is given by,
${v_1} = \sqrt {\dfrac{{{T_1}}}{{{m_1}}}} \,....................\left( 2 \right)$
Then,
${v_2} = \sqrt {\dfrac{{{T_2}}}{{{m_2}}}} \,....................\left( 3 \right)$
By dividing the equation (2) by the equation (3), then the relation between the two velocity is given as,
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\sqrt {\dfrac{{{T_1}}}{{{m_1}}}} }}{{\sqrt {\dfrac{{{T_2}}}{{{m_2}}}} }}$
The mass of the string is same in all the point of the string, then the above equation is written as,
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\sqrt {{T_1}} }}{{\sqrt {{T_2}} }}$
Now, the tension of the string is written as the product of the mass and the acceleration due to gravity, then the above equation is written as,
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\sqrt {2 \times 9.8} }}{{\sqrt {8 \times 9.8} }}$
By cancelling the same terms in the above equation, then the above equation is written as,
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\sqrt 2 }}{{\sqrt 8 }}$
By rearranging the terms in the above equation, then the above equation is written as,
$\dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{2}{8}}$
By cancelling the terms in the above equation, then the above equation is written as,
$\dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{1}{4}}$
By rearranging the terms in the above equation, then the above equation is written as,
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\sqrt 1 }}{{\sqrt 4 }}$
By taking square root in the above equation, then the above equation is written as,
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{1}{2}$
By rearranging the terms in the above equation, then the above equation is written as,
${v_2} = 2{v_1}$
From the above relation of the velocity, then the wavelength relation is given as,
${\lambda _2} = 2{\lambda _1}$
By substituting the wavelength in the above equation, then the above equation is written as,
${\lambda _2} = 2 \times 0.06$
By multiplying the terms in the above equation, then the above equation is written as,
${\lambda _2} = 0.12\,m$

Hence, the option (C) is the correct answer.

Note The velocity of the string is directly proportional to the square root of the tension of the string and inversely proportional to the square root of the mass of the string. As the square root of the tension of the string increases then the velocity of the string increases.