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A string is stretched betweeb fixed poin...

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

`10.5 Hz`

B

`105 Hz`

C

`1.05 Hz`

D

`1050 Hz`

Text Solution

Verified by Experts

The correct Answer is:
2
`(n + 1) (v)/(2l) = 420 …..(1)`
`(nv)/(2l) = 315 ….(2)`
`(1) - (2) (V)/(mu) = 105 Hz , f_(min) = = 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 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
    (a) `105 Hz`
    B
    (b) `1.05 Hz`
    C
    ( c ) `1050 Hz`
    D
    (d) `10.5 Hz`

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

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