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3 years ago

A pipe is open at both ends. The pipe has resonant frequencies of 528 Hz and 660HZ (among others).

1 answer:

7 0

To develop this problem it is necessary to apply the oscillation frequency-related concepts specifically in string or pipe close at both ends or open at both ends.

By definition the oscillation frequency is defined as

$f = n\frac{v}{2L}$

Where

v = speed of sound

L = Length of the pipe

n = any integer which represent the number of repetition of the spectrum (n)1,2,3...)(Number of harmonic)

Re-arrange to find L,

$f = n\frac{v}{2L}\\L = \frac{nv}{2f}$

The radius between the two frequencies would be 4 to 5,

$\frac{528Hz}{660Hz}= \frac{4}{5}$

$4:5$

Therefore the frequencies are in the ratio of natural numbers. That is

$4f = 528\\f = \frac{528}{4}\\f = 132Hz$

Here f represents the fundamental frequency.

Now using the expression to calculate the Length we have

$L = \frac{nv}{2f}\\L = \frac{(1)343m/s}{2(132)}\\L = 1.29m$

Therefore the length of the pipe is 1.3m

For the second harmonic n=2, then

$L = \frac{nv}{2f}\\L = \frac{(2)343m/s}{2(132)}\\L = 2.59m$

Therefore the length of the pipe in the second harmonic is 2.6m

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A 878-kg (1940 lb) dragster, starting from rest, attains a speed of 25.9 m/s (57.9 mph) in 0.62 s. (a) Find the average accelera

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Answer:

41.8m/s^2

Explanation:

Since the dragster starts from rest, initial velocity (u) = 0m/s, final velocity (v) = 25.9m/s, time (t) = 0.62s

From the equations of motion, v = u + at

a = (v - u)/t = (25.9 - 0)/0.62 = 25.9/0.62 = 41.8m/s^2

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3 years ago

A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

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To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

$\sum F_x = 0$

$f-N_w = 0$

$N_w = f$

The forces in the horizontal direction would be,

$\sum F_y = 0$

$N_f -W =0$

$N_f = W$

The sum of Torques at equilibrium,

$\sum \tau = 0$

$Wdcos\theta - N_wLsin\theta = 0$

$WdCos\theta = fLSin\theta$

$f = \frac{Wd}{Ltan\theta}$

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

$f_{max} = \mu W=\frac{Wd}{Ltan\theta}$