MCQ
Answer the following by appropriately matching the lists based on the information given in the paragraph.

A musical instrument is made using four different metal strings, $1,2,3$ and $4$ with mass per unit length $\mu, 2 \mu, 3 \mu$ and $4 \mu$ respectively. The instrument is played by vibrating the strings by varying the free length in between the range $L _0$ and $2 L _0$. It is found that in string-$1$ $(\mu)$ at free length $L _0$ and tension $T _0$ the fundamental mode frequency is $f _0$.

$List-I$ gives the above four strings while $list-II$ lists the magnitude of some quantity.

$List-I$ $List-II$
$(I)$ String-1( $\mu$ ) $(P) 1$
$(II)$ String-2 $(2 \mu)$ $(Q)$ $1 / 2$
$(III)$ String-3 $(3 \mu)$ $(R)$ $1 / \sqrt{2}$
$(IV)$ String-4 $(4 \mu)$ $(S)$ $1 / \sqrt{3}$
  $(T)$ $3 / 16$
  $(U)$ $1 / 16$

($1$) If the tension in each string is $T _0$, the correct match for the highest fundamental frequency in $f _0$ units will be,

$(1)$ $I \rightarrow P , II \rightarrow R , III \rightarrow S , IV \rightarrow Q$

$(2)$ $I \rightarrow P , II \rightarrow Q , III \rightarrow T , IV \rightarrow S$

$(3)$ $I \rightarrow Q , II \rightarrow S , III \rightarrow R , IV \rightarrow P$

$(4)$ I $\rightarrow Q , II \rightarrow P , III \rightarrow R$, IV $\rightarrow T$

($2$) The length of the string $1,2,3$ and 4 are kept fixed at $L _0, \frac{3 L _0}{2}, \frac{5 L _0}{4}$ and $\frac{7 L _0}{4}$, respectively. Strings $1,2,3$ and 4 are vibrated at their $1^{\text {tt }}, 3^{\text {rd }}, 5^{\text {m }}$ and $14^{\star}$ harmonics, respectively such that all the strings have same frequency. The correct match for the tension in the four strings in the units of $T _0$ will be.

$(1)$ $I \rightarrow P , II \rightarrow Q , III \rightarrow T , IV \rightarrow U$

$(2)$ $I \rightarrow T , II \rightarrow Q , III \rightarrow R$, IV $\rightarrow U$

$(3)$ $I \rightarrow P , II \rightarrow Q , III \rightarrow R , IV \rightarrow T$

$(4)$ I $\rightarrow P , II \rightarrow R , III \rightarrow T , IV \rightarrow U$

  • A
    $1,2$
  • B
    $1,3$
  • $1,1$
  • D
    $1,4$

Answer

Correct option: C.
$1,1$
c
For fundamental mode

(image)

$f =\frac{ V }{\lambda}=\frac{1}{2 L } \sqrt{\frac{ T }{\mu}}$

For string $(1)$

$f_0=\frac{1}{2 L} \sqrt{\frac{T}{\mu}} \Rightarrow(P)$

For string $(2)$

$f =\frac{1}{2 L } \sqrt{\frac{ T }{2 \mu}}=\frac{ f _0}{\sqrt{2}} \Rightarrow( R )$

For string $(3)$

$f =\frac{1}{2 L } \sqrt{\frac{ T }{3 \mu}}=\frac{ f _0}{\sqrt{3}} \Rightarrow( S )$

For string $(4)$

$f=\frac{1}{2 L} \sqrt{\frac{T}{4 \mu}}=\frac{f_0}{2} \Rightarrow(Q)$

($2$) For string $(1)$

Length of string $= L _0$

It is vibrating in $I ^{ ts }$ harmonic i.e. fundamental mode.

(image)

$f _0=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}} \Rightarrow( P )$

For string $(2)$

Length of string $=\frac{3 L _0}{2}$

It is vibrating in $III ^{\text {rd }}$ harmonic but frequency is still $f _0$.

$f_0=\frac{3 v}{2 L}$

(image)

$f _0=\frac{3}{2\left(\frac{3 L _0}{2}\right)} \sqrt{\frac{ T _2}{2 \mu}}$

$\Rightarrow f _0=\frac{1}{ L _0} \sqrt{\frac{ T _2}{2 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}}$

$\Rightarrow T _2=\frac{ T _0}{2} \Rightarrow \text { (Q) }$

For string $(3)$

Length of string $=\frac{5 L _0}{4}$

It is vibrating in $5^{\text {th }}$ harmonic but frequency is still $f _0$.

$f _0=\frac{5 V }{2 L }$

(image)

$\Rightarrow f _0=\frac{5}{2\left(\frac{5 L _0}{4}\right)} \sqrt{\frac{ T _3}{3 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}}$

$\Rightarrow \frac{2}{ L _0} \sqrt{\frac{ T _3}{3 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}}$

$T _3=\frac{3 T _0}{16} \Rightarrow( T )$

For string $(4)$

Length of string $=\frac{7 L _0}{4}$

It is vibrating in $14^{\text {th }}$ harmonic but frequency is still $f _0$.

(image)

$f _0=\frac{14 v }{2 L }$

$\begin{array}{l}\Rightarrow \quad f _0=\frac{14}{2\left(\frac{7 L _0}{4}\right)} \sqrt{\frac{ T _4}{4 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}} \\ \Rightarrow \frac{4}{ L _0} \sqrt{\frac{ T _4}{4 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}} \Rightarrow T _4=\frac{ T _0}{16} \Rightarrow( U ) \\\end{array}$

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