Question

In the arrangement shown in figure (15-E6), the string has a mass of 4⋅5 g. How much time will it take for a transverse disturbance produced at the floor to reach the pulley? Take g = 10 m s−2.
Figure

Answer 1

Time Taken will be 0.02 sec

Explanation:

Total length of string 2 + 0.25 = 2.25 mt

Mass per unit length m = $\frac {4.5 \times 10^-^3}{2.25}$

= $2.25\ \times 10^-^3\ kgm^-^1$

T = 2g = 20 N

Wave speed, v = $\sqrt (\frac {T}{m})$

= $\sqrt (\frac {20}{2 \times 10^-^3}$

= $\sqrt 10^4$

= $10^2\ ms^-^1$

= $100\ ms^-^1$

Time taken to reach the pulley, t = $\frac {s}{v}$

= $\frac {2}{100}$ = 0.02 sec

Answer 2

Answer:

Explanation:

Total length of string 2 + 0.25 = 2.25 mt

Mass per unit length m = \frac {4.5 \times 10^-^3}{2.25}

= 2.25\ \times 10^-^3\ kgm^-^1

T = 2g = 20 N

Wave speed, v = \sqrt (\frac {T}{m})

= \sqrt (\frac {20}{2 \times 10^-^3}

= \sqrt 10^4

= 10^2\ ms^-^1

= 100\ ms^-^1

Time taken to reach the pulley, t = \frac {s}{v}

= \frac {2}{100} = 0.02 sec