Question

A long rubber tube having mass $0.9\,{\text{kg}}$ is fastened to a fixed support and the free end of the tube is attached to a cord which passes over a pulley and supports an object, with a mass of $5\,{\text{kg}}$ as shown in fig. If the tube is struck by a transverse blow at one end, the time required for the pulse to reach the other end is:

A.$5\,{\text{s}}$
B.$0.47\,{\text{s}}$
C.$4.7\,{\text{s}}$
D.$3.2\,{\text{s}}$

Answer 1

Use expression for Newton’s second law of motion and determine the tension in the cord. Use the formula for velocity of the wave on a string and determine the velocity of the pulse on the rubber tube. Use the formula for velocity of an object in terms of distance and time and determine the time required for the pulse to reach other end of the tube.

Formulae used:
The expression for Newton’s second law of motion
${F_{net}} = ma$ …… (1)
Here, ${F_{net}}$ is the net force acting on the object, $m$ is mass of the object and $a$ is acceleration of the object.
The velocity $v$ of a wave on the string is given by
$v = \sqrt {\dfrac{T}{\mu }}$ …… (2)
Here, $T$ is tension in the string and $\mu$ is mass per unit length of the string.
The velocity $v$ of an object is
$v = \dfrac{d}{t}$ …… (3)
Here, $d$ is the displacement of the object and $t$ is the time required for displacement.

Complete step by step answer:
We have given that a long rubber tube has mass of $0.9\,{\text{kg}}$ and its free end of the tube is attached to a cord which passes over a pulley and supports an object, with a mass of $5\,{\text{kg}}$ as shown in the figure.
$M = 0.9\,{\text{kg}}$
$m = 5\,{\text{kg}}$

We have asked to determine the time required for the transverse pulse to reach the other end of the tube. From the figure, we can see that the length of the rubber tube is $12\,{\text{m}}$.
$L = 12\,{\text{m}}$
Let us first determine the velocity of the transverse pulse. There is a tension $T$ in the cord at both the ends of which a rubber tube and an object are hanging.
Apply Newton’s second law to the object hanging at the end of the cord.
$T = mg$
Substitute $5\,{\text{kg}}$ for $m$ and $9.8\,{\text{m/}}{{\text{s}}^2}$ for $g$ in the above equation.
$T = \left( {5\,{\text{kg}}} \right)\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)$
$\Rightarrow T = 49\,{\text{N}}$
Hence, the tension in the cord is $49\,{\text{N}}$.

Let us determine mass per unit length of the rubber tube.
$\mu = \dfrac{M}{L}$
Substitute $0.9\,{\text{kg}}$ for $M$ and $12\,{\text{m}}$ for $L$ in the above equation.
$\mu = \dfrac{{0.9\,{\text{kg}}}}{{12\,{\text{m}}}}$
$\Rightarrow \mu = 0.075\,{\text{kg/m}}$
Hence, the mass per unit length of the rubber tube is $0.075\,{\text{kg/m}}$.

Substitute $49\,{\text{N}}$ for $T$ and $0.075\,{\text{kg/m}}$ for $\mu$ equation (2).
$v = \sqrt {\dfrac{{49\,{\text{N}}}}{{0.075\,{\text{kg/m}}}}}$
$\Rightarrow v = \sqrt {653.3}$
$\Rightarrow v = 25.56\,{\text{m/s}}$
Hence, the velocity of the transverse pulse is $25.56\,{\text{m/s}}$.

Let us determine the time required for the pulse to travel the length of the rubber tube and reach the other end. Rearrange equation (3) for time required to reach the other end of the tube.
$t = \dfrac{L}{v}$
Substitute $12\,{\text{m}}$ for $L$ and $25.56\,{\text{m/s}}$ for $v$ in the above equation.
$t = \dfrac{{12\,{\text{m}}}}{{25.56\,{\text{m/s}}}}$
$\therefore t = 0.47\,{\text{s}}$
Therefore, the time required for the pulse to reach the other end of the tube is $0.47\,s$.

Hence, the correct option is B.

Note: The students may think that the tension in the cord over the pulley having a rubber tube and an object at both of its ends is different. But the tension for the cord on both sides of the pulley is the same as the net tension on the cord is due to the resultant force of weight of the rubber tune and the object.