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Montano1993 [528]
3 years ago
7

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

ind a) the fundamental frequency, and b) the speed of the waves on the string.

Physics
1 answer:
Alja [10]3 years ago
4 0

Answer:

\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}

f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}

b) Wave speed

(i) Calculate the wavelength

In a  fundamental vibration, the length of the string is half the wavelength.

\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}

(b) Calculate the speed s

\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}

\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}

\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}

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2 years ago
4. Una cuerda de acero de piano mide 1.60 m de longitud y 0.20 cm de diámetro. ¿Cuál es la tensión en la cuerda si se estira 0.2
bazaltina [42]

Answer:

1030.83\ \text{N}

Explanation:

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T = tensión en cuerda

A = Área de la cadena = \dfrac{\pi}{4}d^2

d = Diámetro de la cuerda = 0.2 cm

L = Longitud original de la cuerda = 1.6 m

El cambio de longitud de una cuerda viene dado por

\Delta L=\dfrac{TL}{AE}\\\Rightarrow T=\dfrac{\Delta LAE}{L}\\\Rightarrow T=\dfrac{0.25\times 10^{-2}\times \dfrac{\pi}{4}(0.2\times 10^{-2})^2\times 210\times 10^9}{1.6}\\\Rightarrow T=1030.84\ \text{N}

La tensión en la cuerda es 1030.84\ \text{N}.

8 0
2 years ago
A 37 cm long solenoid, 1.8 cm in diameter, is to produce a 0.50 T magnetic field at its center. If the maximum current is 4.4 A,
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Answer:

33,458.71 turns

Explanation:

Given: L = 37 cm = 0.37 m, B= 0.50 T, I = 4.4 A, n= number of turn per meter

μ₀ = Permeability of free space = 4 π × 10 ⁻⁷

Solution:

We have B = μ₀ × n × I

⇒ n = B/ (μ₀ × I)

n = 0.50 T / ( 4 π × 10 ⁻⁷ × 4.4 A)

n = 90,428.94 turn/m

No. of turn through 0.37 m long solenoid = 90,428.94 turn/m × 0.37

= 33,458.71 turns

8 0
3 years ago
Read 2 more answers
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