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 L0 and 2L0. It is found that in string-1 \mu at free length L0 and tension T0 the fundamental mode frequency is f0. List-I gives the above four strings while list-II lists the magnitude of some quantity. The length of the strings 1, 2, 3 and 4 are kept fixed at L0, 3L_0 \over 2 , 5L_0 \over 4 and 7L_0 \over 4 respectively. Strings 1, 2, 3 and 4 are vibrated at their 1st, 3rd, 5th and 14th harmonies, respectively such that all the strings have same frequency. The correct match for the tension in the four strings in the units of T0 will be

I \to P, II \to Q, III \to T, IV \to U

I \to P, II \to R, III \to T, IV \to U

I \to P, II \to Q, III \to R, IV \to T

I \to T, II \to Q, III \to R, IV \to U

Waves PYQ — JEE Advanced 2019 · Shift 2 · Q52 with Solution

View written solutionFree

Correct answer: C

Fundamental frequency formula for a stretched string

For a string of free length $L$ , tension $T$ , and linear mass density $\mu$ , the fundamental frequency is

f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

The $n^{\text{th}}$ harmonic frequency is

f_n = n\,f_1 = \frac{n}{2L}\sqrt{\frac{T}{\mu}}

Reference condition from string 1

Given for string-1:

mass per unit length $= \mu$
free length $= L_0$
tension $= T_0$
fundamental frequency $= f_0$

So,

f_0 = \frac{1}{2L_0}\sqrt{\frac{T_0}{\mu}}

This will be our reference relation.

Condition in the question

The four strings are kept at lengths:

String 1: $L_1 = L_0$
String 2: $L_2 = \frac{3L_0}{2}$
String 3: $L_3 = \frac{5L_0}{4}$
String 4: $L_4 = \frac{7L_0}{4}$

Their linear densities are:

String 1: $\mu_1 = \mu$
String 2: $\mu_2 = 2\mu$
String 3: $\mu_3 = 3\mu$
String 4: $\mu_4 = 4\mu$

They vibrate in harmonics:

String 1: 1st harmonic $\Rightarrow n_1=1$
String 2: 3rd harmonic $\Rightarrow n_2=3$
String 3: 5th harmonic $\Rightarrow n_3=5$
String 4: 14th harmonic $\Rightarrow n_4=14$

All have the same frequency.

Since string 1 at 1st harmonic has frequency equal to the common frequency,

f = \frac{1}{2L_0}\sqrt{\frac{T_1}{\mu}}

Matching with the reference form,

f = f_0 \sqrt{\frac{T_1}{T_0}}

Now compute each tension such that all frequencies are equal. The options imply values labeled as $P,Q,R,T,U$ , so we derive the numerical multipliers in units of $T_0$ .

String 1

For string 1,

f = \frac{1}{2L_0}\sqrt{\frac{T_1}{\mu}}

To match the reference frequency setup naturally, take the common frequency as $f_0$ , hence

\frac{1}{2L_0}\sqrt{\frac{T_1}{\mu}} = \frac{1}{2L_0}\sqrt{\frac{T_0}{\mu}}

Therefore,

T_1 = T_0

So string 1 corresponds to multiplier $1$ .

String 2

For string 2:

f = \frac{3}{2L_2}\sqrt{\frac{T_2}{2\mu}}

Substitute $L_2 = \frac{3L_0}{2}$ :

f = \frac{3}{2\cdot \frac{3L_0}{2}}\sqrt{\frac{T_2}{2\mu}} = \frac{1}{L_0}\sqrt{\frac{T_2}{2\mu}}

Set this equal to $f_0$ :

\frac{1}{L_0}\sqrt{\frac{T_2}{2\mu}} = \frac{1}{2L_0}\sqrt{\frac{T_0}{\mu}}

Cancel $\frac{1}{L_0}$ :

\sqrt{\frac{T_2}{2\mu}} = \frac{1}{2}\sqrt{\frac{T_0}{\mu}}

Square both sides:

\frac{T_2}{2\mu} = \frac{1}{4}\cdot \frac{T_0}{\mu}

T_2 = \frac{T_0}{2}

So string 2 corresponds to multiplier $\frac{1}{2}$ .

String 3

For string 3:

f = \frac{5}{2L_3}\sqrt{\frac{T_3}{3\mu}}

Substitute $L_3 = \frac{5L_0}{4}$ :

f = \frac{5}{2\cdot \frac{5L_0}{4}}\sqrt{\frac{T_3}{3\mu}} = \frac{2}{L_0}\sqrt{\frac{T_3}{3\mu}}

Set equal to $f_0$ :

\frac{2}{L_0}\sqrt{\frac{T_3}{3\mu}} = \frac{1}{2L_0}\sqrt{\frac{T_0}{\mu}}

So,

2\sqrt{\frac{T_3}{3\mu}} = \frac{1}{2}\sqrt{\frac{T_0}{\mu}}

\sqrt{\frac{T_3}{3\mu}} = \frac{1}{4}\sqrt{\frac{T_0}{\mu}}

Squaring,

\frac{T_3}{3\mu} = \frac{1}{16}\cdot \frac{T_0}{\mu}

T_3 = \frac{3T_0}{16}

So string 3 corresponds to multiplier $\frac{3}{16}$ .

String 4

For string 4:

f = \frac{14}{2L_4}\sqrt{\frac{T_4}{4\mu}}

Substitute $L_4 = \frac{7L_0}{4}$ :

f = \frac{14}{2\cdot \frac{7L_0}{4}}\sqrt{\frac{T_4}{4\mu}} = \frac{4}{L_0}\sqrt{\frac{T_4}{4\mu}}

Since

\sqrt{\frac{T_4}{4\mu}} = \frac{1}{2}\sqrt{\frac{T_4}{\mu}},

we get

f = \frac{2}{L_0}\sqrt{\frac{T_4}{\mu}}

Set equal to $f_0$ :

\frac{2}{L_0}\sqrt{\frac{T_4}{\mu}} = \frac{1}{2L_0}\sqrt{\frac{T_0}{\mu}}

Thus,

2\sqrt{\frac{T_4}{\mu}} = \frac{1}{2}\sqrt{\frac{T_0}{\mu}}

\sqrt{\frac{T_4}{\mu}} = \frac{1}{4}\sqrt{\frac{T_0}{\mu}}

Squaring,

T_4 = \frac{T_0}{16}

So string 4 corresponds to multiplier $\frac{1}{16}$ .

Final matching

Hence, in units of $T_0$ :

String 1 $\to 1$
String 2 $\to \frac{1}{2}$
String 3 $\to \frac{3}{16}$
String 4 $\to \frac{1}{16}$

Thus the labels must be:

$P = 1$
$Q = \frac{1}{2}$
$T = \frac{3}{16}$
$U = \frac{1}{16}$

So the correct matching is

\text{I} \to P,\quad \text{II} \to Q,\quad \text{III} \to T,\quad \text{IV} \to U

which is Option C.

Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.

More from Waves

View written solutionFree

Correct answer: C

Fundamental frequency formula for a stretched string

For a string of free length $L$ , tension $T$ , and linear mass density $\mu$ , the fundamental frequency is

f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

The $n^{\text{th}}$ harmonic frequency is

f_n = n\,f_1 = \frac{n}{2L}\sqrt{\frac{T}{\mu}}

Reference condition from string 1

Given for string-1:

mass per unit length $= \mu$
free length $= L_0$
tension $= T_0$
fundamental frequency $= f_0$

So,

f_0 = \frac{1}{2L_0}\sqrt{\frac{T_0}{\mu}}

This will be our reference relation.

Condition in the question

The four strings are kept at lengths:

String 1: $L_1 = L_0$
String 2: $L_2 = \frac{3L_0}{2}$
String 3: $L_3 = \frac{5L_0}{4}$
String 4: $L_4 = \frac{7L_0}{4}$

Their linear densities are:

String 1: $\mu_1 = \mu$
String 2: $\mu_2 = 2\mu$
String 3: $\mu_3 = 3\mu$
String 4: $\mu_4 = 4\mu$

They vibrate in harmonics:

String 1: 1st harmonic $\Rightarrow n_1=1$
String 2: 3rd harmonic $\Rightarrow n_2=3$
String 3: 5th harmonic $\Rightarrow n_3=5$
String 4: 14th harmonic $\Rightarrow n_4=14$

All have the same frequency.

Since string 1 at 1st harmonic has frequency equal to the common frequency,

f = \frac{1}{2L_0}\sqrt{\frac{T_1}{\mu}}

Matching with the reference form,

f = f_0 \sqrt{\frac{T_1}{T_0}}

Now compute each tension such that all frequencies are equal. The options imply values labeled as $P,Q,R,T,U$ , so we derive the numerical multipliers in units of $T_0$ .

String 1

For string 1,

f = \frac{1}{2L_0}\sqrt{\frac{T_1}{\mu}}

To match the reference frequency setup naturally, take the common frequency as $f_0$ , hence

\frac{1}{2L_0}\sqrt{\frac{T_1}{\mu}} = \frac{1}{2L_0}\sqrt{\frac{T_0}{\mu}}

Therefore,

T_1 = T_0

So string 1 corresponds to multiplier $1$ .

String 2

For string 2:

f = \frac{3}{2L_2}\sqrt{\frac{T_2}{2\mu}}

Substitute $L_2 = \frac{3L_0}{2}$ :

f = \frac{3}{2\cdot \frac{3L_0}{2}}\sqrt{\frac{T_2}{2\mu}} = \frac{1}{L_0}\sqrt{\frac{T_2}{2\mu}}

Set this equal to $f_0$ :

\frac{1}{L_0}\sqrt{\frac{T_2}{2\mu}} = \frac{1}{2L_0}\sqrt{\frac{T_0}{\mu}}

Cancel $\frac{1}{L_0}$ :

\sqrt{\frac{T_2}{2\mu}} = \frac{1}{2}\sqrt{\frac{T_0}{\mu}}

Square both sides:

\frac{T_2}{2\mu} = \frac{1}{4}\cdot \frac{T_0}{\mu}

T_2 = \frac{T_0}{2}

So string 2 corresponds to multiplier $\frac{1}{2}$ .

String 3

For string 3:

f = \frac{5}{2L_3}\sqrt{\frac{T_3}{3\mu}}

Substitute $L_3 = \frac{5L_0}{4}$ :

f = \frac{5}{2\cdot \frac{5L_0}{4}}\sqrt{\frac{T_3}{3\mu}} = \frac{2}{L_0}\sqrt{\frac{T_3}{3\mu}}

Set equal to $f_0$ :

\frac{2}{L_0}\sqrt{\frac{T_3}{3\mu}} = \frac{1}{2L_0}\sqrt{\frac{T_0}{\mu}}

So,

2\sqrt{\frac{T_3}{3\mu}} = \frac{1}{2}\sqrt{\frac{T_0}{\mu}}

\sqrt{\frac{T_3}{3\mu}} = \frac{1}{4}\sqrt{\frac{T_0}{\mu}}

Squaring,

\frac{T_3}{3\mu} = \frac{1}{16}\cdot \frac{T_0}{\mu}

T_3 = \frac{3T_0}{16}

So string 3 corresponds to multiplier $\frac{3}{16}$ .

String 4

For string 4:

f = \frac{14}{2L_4}\sqrt{\frac{T_4}{4\mu}}

Substitute $L_4 = \frac{7L_0}{4}$ :

f = \frac{14}{2\cdot \frac{7L_0}{4}}\sqrt{\frac{T_4}{4\mu}} = \frac{4}{L_0}\sqrt{\frac{T_4}{4\mu}}

Since

\sqrt{\frac{T_4}{4\mu}} = \frac{1}{2}\sqrt{\frac{T_4}{\mu}},

we get

f = \frac{2}{L_0}\sqrt{\frac{T_4}{\mu}}

Set equal to $f_0$ :

\frac{2}{L_0}\sqrt{\frac{T_4}{\mu}} = \frac{1}{2L_0}\sqrt{\frac{T_0}{\mu}}

Thus,

2\sqrt{\frac{T_4}{\mu}} = \frac{1}{2}\sqrt{\frac{T_0}{\mu}}

\sqrt{\frac{T_4}{\mu}} = \frac{1}{4}\sqrt{\frac{T_0}{\mu}}

Squaring,

T_4 = \frac{T_0}{16}

So string 4 corresponds to multiplier $\frac{1}{16}$ .

Final matching

Hence, in units of $T_0$ :

String 1 $\to 1$
String 2 $\to \frac{1}{2}$
String 3 $\to \frac{3}{16}$
String 4 $\to \frac{1}{16}$

Thus the labels must be:

$P = 1$
$Q = \frac{1}{2}$
$T = \frac{3}{16}$
$U = \frac{1}{16}$

So the correct matching is

\text{I} \to P,\quad \text{II} \to Q,\quad \text{III} \to T,\quad \text{IV} \to U

which is Option C.

Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.

Waves question

Waves question

More from Waves

Waves question

Waves question

More from Waves