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A string is stretched between fixed poin...

A string is stretched between fixed points separated by `75.0cm`. It is observed to have resonant frequencies of `420 Hz` and `315 Hz`. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

A

105 Hz

B

1.05 Hz

C

1050 Hz

D

10.5 Hz

Text Solution

Verified by Experts

The correct Answer is:
A
For string fixed at both the ends, resnant frequency are given by `f = (nv)/(2 L)`. It is given that `315 Hz` and `420 Hz` are two consecutive resonant frequencies. Let these be `nth` and `(n + 1) th` harmonices.
`315 = (nv)/(2 L)` ..(i)
`420 = ((n + 1)v)/(2 L)`...(ii)
Dividing Eq. (i) by Eq. (ii), we get
`(315)/(420) = (n)/( n + 1) rArr n = 3`
Lowest resonant frequency,
`f_0 = (v)/(2 L) = (315)/(3) = 105 Hz`.
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Knowledge Check

  • A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequuencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two . The lowest resonant frequency of this string is

    A
    105 Hz
    B
    155 Hz
    C
    205 Hz
    D
    10.5 Hz

  • A string is stretched betweeb fixed points separated by 75.0 cm . It observed to have resonant frequencies of 420 Hz and 315 Hz . There are no other resonant frequencies between these two. The lowest resonant frequency for this strings is

    A
    `75Hz`
    B
    `110Hz`
    C
    `95Hz`
    D
    `105Hz`

  • A string that is stretched between fixed supports separated by 75.0 cm has resonant frequencies of 420 and 315 Hz, with no intermediate resonant frequencies. What is the lowest resonant frequency?

    A
    250 Hz
    B
    317 Hz
    C
    180 Hz
    D
    105 Hz

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