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Gennadij [26K]
3 years ago
12

When a mass of 5 kg hangs from a vertical wire of length 2 m, waves travel on this wire with a speed of 10 m/s. What will the sp

eed of these waves be, if the mass is doubled without stretching the wire?
14 m/s
20 m/s
5.0 m/s
12 m/s

Physics
2 answers:
ololo11 [35]3 years ago
6 0

Answer:

its 14 m/s

Explanation:

Lena [83]3 years ago
6 0

The speed of these waves will be 14 m/s

\texttt{ }

<h3>Further explanation</h3>

Let's recall the speed of wave formula as follows:

\boxed {v = \lambda \times f}

<em>where:</em>

<em>c = speed of wave ( m/s )</em>

<em>λ = wavelength ( m )</em>

<em>f = frequency of wave ( Hz )</em>

\texttt{ }

\boxed {v = \sqrt{\frac{T}{\mu}}}

<em>where:</em>

<em>v = speed of wave on string ( m/s )</em>

<em>T = tension in the string ( N )</em>

<em>μ = linear density ( kg/m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

initial mass = m₁ = 5 kg

length of vertical wire = L = 2 m

initial speed of wave = v₁ = 10 m/s

final mass = m₂ = 2m₁ = 2(5) = 10 kg

<u>Asked:</u>

final speed of wave = v₂ = ?

<u>Solution:</u>

v_1 : v_2 = \sqrt{\frac{T_1}{\mu}} : \sqrt{\frac{T_2}{\mu}}

v_1 : v_2 = \sqrt{T_1} : \sqrt{T_2}

v_1 : v_2 = \sqrt{w_1} : \sqrt{w_2}

v_1 : v_2 = \sqrt{m_1 g} : \sqrt{m_2 g}

v_1 : v_2 = \sqrt{m_1} : \sqrt{m_2}

10 : v_2 = \sqrt{5} : \sqrt{10}

v_2 = 10\sqrt{2} \texttt{ m/s}

v_2 \approx 14 \texttt{ m/s}

\texttt{ }

<h3>Learn more</h3>
  • Frequency of Sound Wave : brainly.com/question/12441470
  • Frequency of Beats : brainly.com/question/12367463
  • Amplitude of Sound Wave : brainly.com/question/2675676
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Wave

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Alexander Somm

, Consultant, Investor Relations at Novelpharm AG (2015-present)

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Sorry, I had to.

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