Speed of Mechanical Wave on String (Transverse wave) Questions in English

(D) The speed of a transverse wave in a string is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the mass per unit length.
At the lower end of the string, the tension $T$ is provided by the load of mass $M$. Let the mass of the string be $m$ and its length be $L$. The tension at the lower end is $T = Mg$.
When the lift accelerates upwards with $a_1 = 2.1 \,ms^{-2}$, the effective acceleration is $g_{eff1} = g + a_1 = 10 + 2.1 = 12.1 \,ms^{-2}$. The tension is $T_1 = M(g + a_1)$.
The wave speed is $v_1 = \sqrt{\frac{M(g+a_1)}{\mu}} = 88 \,ms^{-1}$.
When the lift accelerates downwards with $a_2 = 1.9 \,ms^{-2}$, the effective acceleration is $g_{eff2} = g - a_2 = 10 - 1.9 = 8.1 \,ms^{-2}$. The tension is $T_2 = M(g - a_2)$.
The wave speed is $v_2 = \sqrt{\frac{M(g-a_2)}{\mu}}$.
Taking the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{g-a_2}{g+a_1}} = \sqrt{\frac{8.1}{12.1}} = \sqrt{\frac{81}{121}} = \frac{9}{11}$.
Therefore, $v_2 = v_1 \times \frac{9}{11} = 88 \times \frac{9}{11} = 8 \times 9 = 72 \,ms^{-1}$.

(D) The speed of a wave on a string is given by $v = \sqrt{T/\mu}$.
Given $T = 500 \ N$,$L_A = 2.5 \ m$,$L_B = 1.5 \ m$,$\mu_A = 2 \times 10^{-4} \ kg/m$,and $\mu_B = 4 \times 10^{-4} \ kg/m$.
The speed in string $A$ is $v_A = \sqrt{500 / (2 \times 10^{-4})} = \sqrt{25 \times 10^5} = 500 \sqrt{2} \ m/s$.
The speed in string $B$ is $v_B = \sqrt{500 / (4 \times 10^{-4})} = \sqrt{12.5 \times 10^5} = 500 \sqrt{0.5} \ m/s$.
The time taken for the pulse to reach the joint is $t = L/v$.
$t_1 = L_A / v_A = 2.5 / (500 \sqrt{2}) = 0.005 / \sqrt{2} \ s$.
$t_2 = L_B / v_B = 1.5 / (500 \sqrt{0.5}) = 0.003 / \sqrt{0.5} \ s$.
Therefore,the ratio $t_1 / t_2 = (0.005 / \sqrt{2}) / (0.003 / \sqrt{0.5}) = (5/3) \times \sqrt{0.5/2} = (5/3) \times \sqrt{0.25} = (5/3) \times 0.5 = 2.5 / 3 \approx 0.833$.
Wait,re-calculating: $t_1/t_2 = (L_A/v_A) / (L_B/v_B) = (L_A/L_B) \times \sqrt{\mu_A/\mu_B} = (2.5/1.5) \times \sqrt{(2 \times 10^{-4}) / (4 \times 10^{-4})} = (5/3) \times \sqrt{0.5} = 1.666 \times 0.707 \approx 1.178 \approx 1.18$.