
The velocity of a transverse wave in a stretched wire is $100m{s^{ - 1}}$. If the length of wire is doubled and tension in the string is also doubled, the final velocity of the transverse wave in the wire is
A. $100m{s^{ - 1}}$
B. $141.4m{s^{ - 1}}$
C. $200m{s^{ - 1}}$
D. $282.8m{s^{ - 1}}$
Answer
580.2k+ views
Hint: The velocity of a transverse wave is equal to the square root of the tension divided by mass per unit length of the wire. The velocity depends on the tension of the wire.
Complete step by step answer: A transverse wave is a type of mechanical wave. A transverse wave motion is that wave motion in which the individual particles of the medium (wire) execute simple harmonic motion about their mean positions in a direction perpendicular to the direction of propagation of the wave. The velocity (v) of propagation of a transverse wave over a wire in a stretched wire is given by the equation $v = \sqrt {\dfrac{T}{m}} $ where T is the tension in the wire and m is known as linear density i.e., mass per unit length of the wire. Let the initial length of the wire be l.
Initially, $v = \sqrt {\dfrac{T}{m}} $
$100 = \sqrt {\dfrac{T}{m}} $ -----[eqn1]
Now, when the length of the wire is doubled i.e., new length be $L = 2l$ and tension in the string gets also doubled i.e., new tension $\left( {{T_1}} \right) = 2T$.
Let $m_1$ be the new mass per unit length of the wire and new velocity be ${v_1}$. The new mass per unit length of the wire remains the same even if its length becomes double because we consider mass per unit length which doesn’t change. Thus, ${m_1} = m$.
Now, ${v_1} = \sqrt {\dfrac{{{T_1}}}{m}} $
$\Rightarrow {v_1} = \sqrt {\dfrac{{2T}}{m}} $
$\Rightarrow {v_1} = \sqrt 2 \times \sqrt {\dfrac{T}{m}} $
$\Rightarrow {v_1} = \sqrt 2 \times 100$ [ ---from eqn1]
$\Rightarrow {v_1} = 1.414 \times 100 = 141.4m{s^{ - 1}}$
Therefore, option B is correct.
Note:The linear density i.e., mass per unit length of the stretched wire remains the same as it means mass of a unit length of the wire which is fixed for a particular wire. The velocity of the transverse wave is directly proportional to the tension and inversely proportional to the mass per unit length of the wire.
Complete step by step answer: A transverse wave is a type of mechanical wave. A transverse wave motion is that wave motion in which the individual particles of the medium (wire) execute simple harmonic motion about their mean positions in a direction perpendicular to the direction of propagation of the wave. The velocity (v) of propagation of a transverse wave over a wire in a stretched wire is given by the equation $v = \sqrt {\dfrac{T}{m}} $ where T is the tension in the wire and m is known as linear density i.e., mass per unit length of the wire. Let the initial length of the wire be l.
Initially, $v = \sqrt {\dfrac{T}{m}} $
$100 = \sqrt {\dfrac{T}{m}} $ -----[eqn1]
Now, when the length of the wire is doubled i.e., new length be $L = 2l$ and tension in the string gets also doubled i.e., new tension $\left( {{T_1}} \right) = 2T$.
Let $m_1$ be the new mass per unit length of the wire and new velocity be ${v_1}$. The new mass per unit length of the wire remains the same even if its length becomes double because we consider mass per unit length which doesn’t change. Thus, ${m_1} = m$.
Now, ${v_1} = \sqrt {\dfrac{{{T_1}}}{m}} $
$\Rightarrow {v_1} = \sqrt {\dfrac{{2T}}{m}} $
$\Rightarrow {v_1} = \sqrt 2 \times \sqrt {\dfrac{T}{m}} $
$\Rightarrow {v_1} = \sqrt 2 \times 100$ [ ---from eqn1]
$\Rightarrow {v_1} = 1.414 \times 100 = 141.4m{s^{ - 1}}$
Therefore, option B is correct.
Note:The linear density i.e., mass per unit length of the stretched wire remains the same as it means mass of a unit length of the wire which is fixed for a particular wire. The velocity of the transverse wave is directly proportional to the tension and inversely proportional to the mass per unit length of the wire.
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