91Ó°ÊÓ

To find the wave speed use the relation between string's tension and mass per unit length. The mass is calculated by using the relation between the frequency and the tension.

The frequency of the oscillator is \({f_n} = 60{\rm{ Hz}}\).

The string’s mass per unit length is \(\mu = 3.5 \times {10^{ - 4}}{\rm{ kg/m}}\).

The length of the vibrating string is \(l = 1.50{\rm{ m}}\).

The mass is \(m\).

The mass hung to create one loop of standing wave is calculated as,

\(\begin{aligned}{c}{f_n} = \frac{n}{{2l}}\sqrt {\frac{F}{\mu }} \\{f_n} = \frac{n}{{2l}}\sqrt {\frac{{{m_1}g}}{\mu }} \\{m_1} = \frac{{4{l^2}f_n^2\mu }}{{{n^2}g}}\end{aligned}\)

Here, F is the tension force, n is the number of overtone and g is the gravitational acceleration.

Substitute the known values to the above equation.

\(\begin{aligned}{c}{m_1} &= \frac{{4 \times {{\left( {1.50{\rm{ m}}} \right)}^2} \times {{\left( {60{\rm{ Hz}}} \right)}^2} \times \left( {3.5 \times {{10}^{ - 4}}{\rm{ kg/m}}} \right)}}{{{{\left( 1 \right)}^2}\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}\\ &= 1.157{\rm{ kg}}\end{aligned}\)

Thus, the mass needed to create one loop is \(1.157{\rm{ kg}}\).

The mass hung to produce two loops of standing waves is given by,

\({m_2} = \frac{{4{l^2}f_n^2\mu }}{{{{\left( n \right)}^2}g}}\)

Here, the value of n for two loops is 2.

Substitute the known values in the above equation.

\(\begin{aligned}{c}{m_2} &= \frac{{4 \times {{\left( {1.50{\rm{ m}}} \right)}^2} \times {{\left( {60{\rm{ Hz}}} \right)}^2} \times \left( {3.5 \times {{10}^{ - 4}}{\rm{ kg/m}}} \right)}}{{{{\left( 2 \right)}^2} \times \left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}\\ &= 0.289{\rm{ kg}}\end{aligned}\)

Thus, the mass required to create two loops is \(0.289{\rm{ kg}}\).

The mass hung to produce five loops of standing waves is,

\({m_5} = \frac{{4{l^2}f_n^2\mu }}{{{{\left( n \right)}^2}g}}\)

Here, the value of n for five loops is 5.

Substitute the known values in the above equation.

\(\begin{aligned}{c}{m_5} &= \frac{{4 \times {{\left( {1.50{\rm{ m}}} \right)}^2} \times {{\left( {60{\rm{ Hz}}} \right)}^2} \times \left( {3.5 \times {{10}^{ - 4}}{\rm{ kg/m}}} \right)}}{{{{\left( 5 \right)}^2} \times \left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}\\ &= 4.62 \times {10^{ - 2}}{\rm{ kg}}\end{aligned}\)

The mass hung to produce five loops is \(4.62 \times {10^{ - 2}}{\rm{ kg}}\).