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Question: The time taken by pulse to reach the other end of rope is: A) \(\sqrt {\dfrac{{2L}}{g}} \). B) \...

The time taken by pulse to reach the other end of rope is:
A) 2Lg\sqrt {\dfrac{{2L}}{g}} .
B) Lg\sqrt {\dfrac{L}{g}} .
C) 2Lg2\sqrt {\dfrac{L}{g}} .
D) L2g\sqrt {\dfrac{L}{{2g}}} .

Explanation

Solution

The fundamental vibration of a string has λ=2L\lambda = 2L also v=λfv = \lambda f where, the velocity is vv, λ\lambda is the wavelength and ff is frequency of the pulse also the rope will have some tension which will be equal to the weight of rope as there is no other mass on the rope. Also the relation between the velocity of pulse and tension of the rope which can be used in solving this problem.

Formula used: The velocity of a pulse on the rope of mass MM, length LL and tension TT is given by v=TLMv = \sqrt {\dfrac{{T \cdot L}}{M}} . Also the velocity of a pulse is given by v=λfv = \lambda \cdot f where vv is the velocity of pulse, λ\lambda is the wavelength and ff is the frequency of the pulse.

Step by step solution:
Step 1.
The velocity of a pulse under tension TT, mass of rope MM and length of rope as LL is given by v=TLMv = \sqrt {\dfrac{{T \cdot L}}{M}} .
Step 2.
Since v=λfv = \lambda f and for fundamental vibration of a string λ=2L\lambda = 2L and velocity is v=TLMv = \sqrt {\dfrac{{T \cdot L}}{M}} .
Therefore,
v=TLM λf=TLM 2Lf=TLM f=12LTLM f=12TLM  v = \sqrt {\dfrac{{T \cdot L}}{M}} \\\ \lambda \cdot f = \sqrt {\dfrac{{T \cdot L}}{M}} \\\ 2L \cdot f = \sqrt {\dfrac{{T \cdot L}}{M}} \\\ f = \dfrac{1}{{2L}} \cdot \sqrt {\dfrac{{T \cdot L}}{M}} \\\ f = \dfrac{1}{2} \cdot \sqrt {\dfrac{T}{{L \cdot M}}} \\\ ………eq.(1)
Step 3.
Since the tension in the string has to be equal to the weight of the body T=MgT = M \cdot g.
Replace the value of T=MgT = M \cdot g in the equation (1).
f=12TLM f=12MgLM f=12gL  f = \dfrac{1}{2} \cdot \sqrt {\dfrac{T}{{L \cdot M}}} \\\ f = \dfrac{1}{2} \cdot \sqrt {\dfrac{{M \cdot g}}{{L \cdot M}}} \\\ f = \dfrac{1}{2} \cdot \sqrt {\dfrac{g}{L}} \\\ ………eq.(2)
Step 4.
As we know that the frequency is the inverse of the time therefore taking the inverse of frequency from equation (2).
f=12gL T=2Lg  f = \dfrac{1}{2} \cdot \sqrt {\dfrac{g}{L}} \\\ T = 2 \cdot \sqrt {\dfrac{L}{g}} \\\
So, the time taken by the pulse to reach to the other end of the rope is given by T=2LgT = 2 \cdot \sqrt {\dfrac{L}{g}} .

So the correct answer for this problem is option C.

Additional information: A wave of wavelength λ\lambda and velocity vv and time taken as TT is given as speed = distancetime v=λt v=λf  {\text{speed = }}\dfrac{{{\text{distance}}}}{{{\text{time}}}} \\\ v = \dfrac{\lambda }{t} \\\ v = \lambda \cdot f \\\
As we know that the inverse of the time is nothing but frequency ff.

Note: Students should remember the formula of the velocity for fundamental vibration of a string also the formula which can be used for similar problems is the relation between the velocity of the pulse and the tension of the string. The tension in the string will be equal to the weight of the rope as there is no mass except the mass of the rope and there are only two forces which are in opposite directions.