2. A 3.5 m long string is fixed at both ends and vibrates in its 7th harmonic with an amplitude (at an antinode) of 2.4 cm. If the speed of

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2. A 3.5 m long string is fixed at both ends and vibrates in its 7th harmonic with an amplitude (at an antinode) of 2.4 cm. If the speed of waves on the string is 150 m/s, what is the maximum speed for a point on the string at an antinode?

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Thiên Hương 3 years 2021-07-31T15:22:43+00:00 1 Answers 16 views 0

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  1. thongdat2
    0
    2021-07-31T15:24:17+00:00
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    Answer:

    Maximum speed for a point on the string at anti node will be 22.6 m/sec

    Explanation:

    We have given length of string L = 3.5 m

    For 7th harmonic length of the string L=\frac{7\lambda }{2}

    So \lambda =\frac{2L}{7}

    Speed of the wave in the string is 150 m/sec

    Frequency corresponding to this wavelength f=\frac{v}{\lambda }=\frac{7v}{2L}

    So angular frequency will be equal to \omega =2\pi f=2\pi \times \frac{7v}{2L}=2\times 3.14\times \frac{7\times 150}{2\times 3.5}=942rad/sec

    Maximum speed is equal to v_m=A\omega =0.024\times 942=22.60m/sec

    So maximum speed for a point on the string at anti node will be 22.6 m/sec

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