Now the friends are ready to tackle a homework problem. A pulse is sent traveling along a rope under a tension of 29 N whose mass per unit l

Question

Now the friends are ready to tackle a homework problem. A pulse is sent traveling along a rope under a tension of 29 N whose mass per unit length abruptly changes, from 19 kg/m to 45 kg/m. The length of the rope is 2.5 m for the first section and 2.8 m for the second, and the second rope is rigidly fixed to a wall. Two pulses will eventually be detected at the origin: the pulse that was reflected from the medium discontinuity and the pulse that was originally transmitted, which hits the wall and is reflected back and transmitted through the first rope. What is the time difference, Δt, between the two pulses detected at the origin? s

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Dâu 4 years 2021-08-07T21:56:29+00:00 1 Answers 58 views 0

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  1. minhkhang
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    2021-08-07T21:58:28+00:00
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    Answer:

    The time difference is 2.97 sec.

    Explanation:

    Given that,

    Tension = 29 N

    Mass per unit length \mu_{1}=19\ kg/m

    Mass per unit length \mu_{2}=45\ kg/m

    Length of first section = 2.5 m

    Length of second section = 2.8 m

    We need to total distance of first pulse

    Using formula for distance

    d=2.5+2.5

    d_{1}=5.0\ m

    We need to total distance of second pulse

    Using formula for distance

    d=2.8+2.8

    d_{2}=5.6\ m

    We need to calculate the speed of pulse in the first string

    Using formula of speed

    v_{1}=\sqrt{\dfrac{T}{\mu_{1}}}

    Put the value into the formula

    v_{1}=\sqrt{\dfrac{29}{19}}

    v_{1}=1.24\ m/s

    We need to calculate the speed of pulse in the second string

    Using formula of speed

    v_{2}=\sqrt{\dfrac{T}}{\mu_{2}}}

    Put the value into the formula

    v_{2}=\sqrt{\dfrac{29}{45}}

    v_{2}=0.80\ m/s

    We need to calculate the time for first pulse

    Using formula of time

    t_{1}=\dfrac{d_{1}}{v_{1}}

    Put the value into the formula

    t_{1}=\dfrac{5.0}{1.24}

    t_{1}=4.03\ sec

    We need to calculate the time for second pulse

    Using formula of time

    t_{2}=\dfrac{d_{1}}{v_{1}}

    Put the value into the formula

    t_{2}=\dfrac{5.6}{0.80}

    t_{2}=7\ sec

    We need to calculate the time difference

    Using formula of time difference

    \Delta t=t_{2}-t_{1}

    Put the value into the formula

    \Delta t=7-4.03

    \Delta t=2.97\ sec

    Hence, The time difference is 2.97 sec.

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