Question

Two identical strings of length I and 2I vibrate with fundamental frequencies N Hz and 1.5 N Hz, respectively. The ratio of tensions for smaller length to large length is ____________.

Options

• 1 : 3

• 1 : 9

• 3 : 1

• 9 : 1

Answer 1

Two identical strings of length I and 2I vibrate with fundamental frequencies N Hz and 1.5 N Hz, respectively. The ratio of tensions for smaller length to large length is 1 : 9.

Explanation:

The fundamental frequency of wave produced in a string is

f = `"n"/(2"l") sqrt("T"/"m")`

where, n = mode of harmonic.

I = length of string,

T = tension in string,

and m = mass per unit length of string.

`"f" ∝ sqrt("T")/"l"`

or `"f"_1/"f"_2 = sqrt("T"_1/"T"_2) xx "l"_2/"l"_1`

Here, f₁ = N, f₂ = 1.5 N, l₁ = I and l₂ = 2l

∴ `"N"/(1.5 "N") = sqrt("T"_1/"T"_2) xx (2"l")/"l"`

`sqrt("T"_1/"T"_2) = 1/3` or 1 : 3

∴ T₁ : T₂ = 1 : 9