Two strings (A, B) having linear densities mu | vedclass

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(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 	imes 10^{-4} \

Vedclass · Accepted Answer

$1.18$

Vedclass · Answer

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

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