# A string is stretched between fixed points separated by 75.0 cm. It is 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 string is: Option 1) 205 Hz Option 2) 10.5 Hz Option 3) 105 Hz Option 4) 155 Hz

As we discussed in

Fundamental frequency with end correction -

$\nu _{0}= \frac{V}{4\left ( l+e \right )}$     (one end open)

$\nu _{0}= \frac{V}{2\left ( l+2e \right )}$    (Both end open)

e = end correction

-

In a stretched string, all multiples of frequencies can be obtained, i.e., If fundamental frequency is $\eta$, then higher frequencies will be $2 \eta, 3\eta, 4\eta\:.......$

According to question,

$\eta =420-315=105\:Hz$

Option 1)

205 Hz

This option is incorrect.

Option 2)

10.5 Hz

This option is incorrect.

Option 3)

105 Hz

This option is correct.

Option 4)

155 Hz

This option is incorrect.

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