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konstantin123 [22]

3 years ago

15

A 1.50-m-long rope is stretched between two supports with a tension that makes the speed of transverse waves 48.0 m/s. What are

the wavelength and frequency of (a) the fundamental tone; (b) the second overtone; (c) the fourth harmonic?

1 answer:

astra-53 [7]3 years ago

4 0

Answer: a) 16Hz, 3m b) 48Hz, 1mc) 80Hz, 0.6m

Explanation:

a) Fundamental frequency in string is represented as Fo = V/2L where;

Fo is the fundamental frequency

V is the speed of the transverse wave = 48m/s

L is the length of the wire. = 1.50m

Substituting this values in the formula given we have;

Fo = 48/2(1.5)

Fo = 48/3

Fo = 16Hz

The fundamental tone is therefore 16Hz

Using v =f¶

Where f is the frequency and ¶ is the wavelength, the wavelength of the fundamental note will be;

¶ = v/fo

¶ = 48/16 = 3m

b) Overtones or harmonics is the multiple integral of the fundamental frequency. The multiples are I'm arithmetical progression.

First overtone f1 = 2fo

Second overtone f2 = 3fo etc.

Since fo = 16Hz

Second overtone f2 = 3×16 = 48Hz

¶ = v/f2 = 48/48

¶ = 1m

c) Fourth harmonic or overtone will be f4 = 5fo

F4 = 5×16 = 80Hz

The fourth harmonic is therefore 80Hz

¶ = v/f4 = 48/80

¶ = 0.6m

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A rock is lifted by a machine to a height of 10m. If it has a mass of 22 kilograms

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3 years ago

A child goes down the slide,starting from rest. If the length of the slide is 2m and it takes the child 3 seconds to go down the

lapo4ka [179]

Answer:

$0.44 m/s^2$

Explanation:

We have the following data:

- distance covered by the child: d = 2 m (length of the slide)

- time taken to cover this distance: t = 3 s

- initial velocity of the child: 0 m/s (he starts from rest)

So we can find the acceleration by using the equation:

$d=ut+\frac{1}{2}at^2$

Where a is the acceleration.

Substituting the values and solving for a,

$a=\frac{2d}{t^2}=\frac{2(2)}{3^2}=0.44 m/s^2$

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3 years ago

Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

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3 years ago