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2 years ago

A bicycle wheel rotates uniformly through 2.0 revolutions in 4.0 s. what is the frequency of the wheel's rotation?

2 answers:

muminat2 years ago

8 0

The frequency is defined as the number of cycles done in one second:
 $f= \frac{N}{t}$ 
where N is the number of cycles and t is the time taken to complete N cycles. The wheel in this problem does N=2 revolutions in a time of t=4.0s, therefore its frequency of rotation is:
 $f= \frac{N}{t} = \frac{2.0}{4.0s} = 0.5 Hz$

Alex17521 [72]2 years ago

8 0

Answer: The frequency of the wheel's rotation is 0.5 Hz.

Explanation:

Frequency of rotation: It is defined as the number of revolution in a second.

The expression for the frequency in terms of the number of revolution in a given time is as follows;

$f=\frac{N}{t}$

Here N is the number of cycles and t is the time taken.

Put N= 2 rev and t= 4 s in the above expression.

$f=\frac{2}{4}$

$f=0.5 Hz$

Therefore, the correct answer is 0.5 Hz.

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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

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

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

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

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

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

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

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

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

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