Waves Ques 40
40. A string is stretched between two 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 :
[2015 RS]
(a) $205 $ $Hz$
(b) $10.5 $ $Hz$
(c) $105 $ $Hz$
(d) $155 $ $Hz$
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Answer:
Correct Answer: 40.(c)
Solution:
- (c) Resonant frequencies, for a string fixed at both ends, will be
$f_n=\frac{n \nu}{2 L}$ where, $n=1,2,3 \ldots$.
The difference between two consecutive resonant frequency is,
$ A _{f n}=f _{n+1}-f_n \Rightarrow \frac{(n+1) v}{2 L}-\frac{n v}{2 L}=\frac{v}{2 L} $
which is also the lowest resonant frequency ( $n=1$ ).
Thus, for the given string the lowest resonant frequency will be
$=420 $ $Hz-315 $ $Hz$
$=105$ $ Hz$
The difference between any two successive frequencies will be ’ $n$ '
According to question, $n=420-315=105$ $ Hz$ So the lowest frequency of the string is $105 $ $Hz$.
In a stretched string all multiples of frequencies can be obtained i.e., if fundamental frequency is $n$ then higher frequencies will be $2 n, 3 n, 4 n$…
BETA
