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2 years ago

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Consider a string of total length L, made up of three segments of equal length. The mass per unit length of the first segment is

μ, that of the second is 2μ, and that of the third μ/4. The third segment is tied to a wall, and the string is stretched by a force of magnitude Ts applied to the first segment; Ts is much greater than the total weight of the string. How long will it take a transverse wave to propagate from one end of the string to the other? Express the time t in terms of L, μ, and Ts.

1 answer:

zzz [600]2 years ago

3 0

Answer:

Explanation:

Length of each segment is $\frac{L}{3}$

Speed of wave in first segment is $v_1=\sqrt{\frac{T_s}{\mu}}$

Speed of wave in second segment is $v_2=\sqrt{\frac{T_s}{2\mu}}$

Speed of wave in third segment is $v_3=\sqrt{\frac{T_s}{\frac{\mu}{4}}}=\sqrt{\frac{4T_s}{\mu}}$

Now time for the transverse wave to propagate is

$t=t_1 + t_2 + t_3\\=\frac{(\frac{L}{3})}{v_1}+\frac{(\frac{L}{3})}{v_2} + \frac{(\frac{L}{3})}{v_3}\\\\=(\frac{L}{3})(\frac{1}{\sqrt{\frac{T_s}{\mu}}} + \frac{1}{\sqrt{\frac{T_s}{2\mu}}} + \frac{1}{\sqrt{\frac{4T_s}{\mu}}})$

simplifying we get

$t=(\frac{3+2\sqrt{2}}{6})L\sqrt{\frac{\mu}{T_s}}$

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0

$1.67\times 10^{-8}\ s$

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t = Time taken

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$F=qE$

Acceleration is given by

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Magnitude of electric field is 2630250 N/C

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The angle counterclockwise from left is zero.

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2 years ago