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 μ ,   2 μ ,   3 μ and 4 μ respectively.…

Question

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

μ, 2 μ, 3 μ

and

4 μ

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 (μ)

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
$(a)$	String $- 1 (μ)$	$(p)$	$1$
$(b)$	String $- 2 (2 μ)$	$(q)$	$\frac{1}{2}$
$(c)$	String $- 3 (3 μ)$	$(r)$	$\frac{1}{\sqrt{2}}$
$(d)$	String $- 4 (4 μ)$	$(s)$	$\frac{1}{\sqrt{3}}$
		$(t)$	$\frac{3}{16}$
		$(u)$	$\frac{1}{16}$

If the tension in each string is

T_{0},

the correct match for the highest fundamental frequency in

f_{0}

units will be,

Aicharya · Accepted Answer

Fundamental frequency is maximum when length is minimum i.e.

L_{0},

Case

1 . L = L_{0}, T = T_{0}, f = f_{0}; f_{1} = \frac{1}{2 L_{0}} \sqrt{\frac{T_{0}}{μ}} = f_{0}

(fundamental frequency of vibration in string fixed at both ends)
Case

2 . f_{2} = \frac{1}{L_{0}} \sqrt{\frac{T_{2}}{2 μ}} = \frac{f_{0}}{\sqrt{2}}

Case

3 . f_{3} = \frac{1}{L_{0}} \sqrt{\frac{T_{2}}{3 μ}} = \frac{f_{0}}{\sqrt{3}}

Case

4 . f_{4} = \frac{1}{L_{0}} \sqrt{\frac{T_{2}}{4 μ}} = \frac{f_{0}}{2}

`

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

Solution