# A string vibrates in three loops under tension produced by a mass weighing 8 g. How much weight...

## Question:

A string vibrates in three loops under tension produced by a mass weighing 8 g. How much weight must be applied so that string vibrates in six loops? Neglect mass of the string.

## Laws of Vibrating String

1) Fundamental frequency of a vibrating is inversely proportional to length of the string, if tension in the string and mass per unit length of the string are kept constant. 2) Fundamental frequency of a vibrating is directly proportional to square root of tension in the string if length of the string and mass per unit length of the string are kept constant.3) Fundamental frequency of a vibrating is inversely proportional to square root of mass per unit length of the string if tension in the string and length of the string are kept constant.

{eq}\displaystyle{ \\ }{/eq}

Symbols Used :-

1) {eq}\space T_1, \space T_2,\space p_1, \space p_2 {/eq} are the initial and final tensions and initial and final number of loops respectively

2) {eq}\space n, \space p,\space L, \space T, \space m {/eq} are the fundamental frequency, length of the string , tension in the string and mass per unit length respectively.

Given Data :-

• {eq}p_1 = 3,\\ p_2 = 6, \\ T_1 =8 \ \rm gwt {/eq}

Required

Tension in the string required for 6 loops =?

Solution:-

The fundamental frequency of a stretched string is given by

\begin{align} n &= \frac{p}{2 \times L} \times \sqrt{ \frac {T}{m}} \\[0.3 cm] \end{align}

If n,L m are kept constant then {eq}p \times \sqrt {T} = \space{/eq} constant

\begin{align} T_1 \times p_1 ^{2} &=T_2 \times p_2^{2} \\[0.3 cm] T_2 &= \frac{ T_1 \times p_1 ^{2}} {p_2^{2}} \\[0.3 cm] T_2 &= \frac{ 8 \ \rm g wt \times 3 ^{2}} {6^{2}} \\[0.3 cm] T_2 &= \frac{ 8 \ \rm g wt \times 9} {36} \\[0.3 cm] T_2 &= \frac{ 8 \ \rm g wt } {4} \\[0.3 cm] T_2 &=2 \ \rm g wt \\[0.3 cm] \end{align}

Tension in the string required for 6 loops is 2 g wt or mass required is 2 gram.