Question

Two long strings A and B, each having linear mass density $1.2 \times {10}^{-2}kg{m}^{-1}$, are stretched by different tensions 4.8N and 7.5N respectively and are kept parallel to each other with their left ends at x=0. Wave pulses are produced on the strings at the left ends at t=0 on string A and t=20 on string B. When and where will the pulse on B overtake that on A?

Answer 1

Hint: To solve this problem, first find the speed of waves on string A and string B using the formula giving relation between speed, tension and the mass. Then, subtract these speeds to get the relative speed between the wave in the string A and wave in the string B. Find the difference between the time for the wave to be produced on string A and string B. Then, find the distance travelled by the wave in that obtained time in string A. Dividing this obtained distance by the calculated relative speed will give the time at which pulse B overtake pulse A.

Complete answer:
Linear mass density of string A (${m}_{A}$)= $1.2 \times {10}^{-2}kg{m}^{-1}$
            Linear mass density of string B (${m}_{B}$)= $1.2 \times {10}^{-2}kg{m}^{-1}$
            Tension on string A (${T}_{A}$)= 4.8N
            Tension on string B (${T}_{B}$)= 7.5N
Let the speed of the waves in string A and B be ${v}_{A}$ and ${v}_{B}$
Relation between speed of the wave and tension on the spring is given by,
$v= \sqrt {\dfrac {T}{m}}$
Where, v is the speed of the wave
            T is tension on the string
            M is the mass of the string
Thus, the speed of the waves in string A is given by,
${v}_{A} =\sqrt {\dfrac {{T}_{A}}{{m}_{A}}}$
Substituting the values in above equation we get,
${v}_{A} =\sqrt {\dfrac {4.8}{1.2 \times {10}^{-2}kg{m}^{-1}}}$
$\Rightarrow {v}_{A}= 20 {m}/{s}$
Similarly, the speed of the waves in string B is given by,
${v}_{B} =\sqrt {\dfrac {{T}_{B}}{{m}_{B}}}$
Substituting the values in above equation we get,
${v}_{B} =\sqrt {\dfrac {7.5}{1.2 \times {10}^{-2}kg{m}^{-1}}}$
$\Rightarrow {v}_{A}= 25 {m}/{s}$
Relative speed between the wave in string A and the wave in string B is given by,
${v}_{0} = {v}_{B}- {v}_{A}$
$\Rightarrow {v}^{0} = 25-20$
$\Rightarrow {v}^{0} = 5 {m}/{s}$
Difference between the time for the wave to be produced on string A and string B is given by,
$ t= 20- 0$
$\Rightarrow t = 20 ms$
$\Rightarrow t = 0.02s$
Distance travelled by the wave in 0.02 s in string A is given by,
$Speed = \dfrac {Distance}{Time}$
Substituting the values in above equation we get,
$20= d \times 0.02$
$\Rightarrow d = 20 \times 0.02$
$\Rightarrow d= 0.4m$
Time taken by the wave in string B to overtake the wave in string A is given by,
$ {v}_{0} = \dfrac {s}{{t}_{0}}$
Substituting values in above equation we get,
$ 5 = \dfrac {0.4}{{t}_{0}}$
$\Rightarrow {t}_{0} = \dfrac {0.4}{5}$
$\Rightarrow {t}_{0}= 0.08s$
Hence, the pulse on string B will overtake the pulse on string A at 0.4m in 0.08s.

Note:
Students should take care while solving this problem that they do not miss any step or any part of the solution. If you miss out any part of the solution then you won’t be able to reach the final solution. From the above solution, it can be understood that the speed of the wave on the string is dependent on the tension on the string and mass per unit length of the string. Speed of the wave on the spring is directly proportional to the square root of the tension.