# Movimiento Ondulatorio 01

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• 1.

### Sea la ecuación de una onda: Y = 3.Cos2π(10t - 0.05x). La frecuencia de esta onda es:

• A.

0.05 Hz

• B.

0.1 Hz

• C.

10 Hz

• D.

20 Hz

• E.

20.π Hz

C. 10 Hz
Explanation
The equation of the wave is given as Y = 3.Cos2π(10t - 0.05x). The frequency of a wave is determined by the coefficient in front of the t term. In this case, the coefficient is 10, which means the frequency of the wave is 10 Hz.

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• 2.

### Sea la ecuación de una onda: Y = 3.Cos2π(10t - 0.05x). La longitud de onda es:

• A.

0.05 cm

• B.

0.1 cm

• C.

10 cm

• D.

20 cm

• E.

20.π cm

D. 20 cm
Explanation
The equation of a wave is given as Y = 3.Cos2π(10t - 0.05x). In this equation, the coefficient of x is -0.05, which represents the wavelength. The wavelength is the distance between two consecutive points in a wave that are in phase. Since the coefficient of x is -0.05, the wavelength is equal to the reciprocal of this coefficient, which is 1/(-0.05) = -20. This negative sign can be ignored as it only indicates the direction of the wave. Therefore, the wavelength is 20 cm.

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• 3.

### Sea la ecuación de una onda: Y = 3.Cos2π(10t - 0.05x). La velocidad de propagación de la onda es:

• A.

0.5 cm / seg

• B.

10 cm / seg

• C.

20 cm / seg

• D.

100 cm / seg

• E.

200 cm / seg

E. 200 cm / seg
Explanation
The equation of a wave is given as Y = 3.Cos2π(10t - 0.05x). The velocity of propagation of a wave is determined by the coefficient of x in the equation. In this case, the coefficient of x is -0.05. The velocity of propagation is calculated by taking the absolute value of this coefficient, which is 0.05. Since the velocity is measured in cm/sec, the answer is 0.05 cm/sec.

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• 4.

### En una cuerda, una onda de amplitud 3 cm, de período π segundos y de longitud de onda 2π cm avanza en sentido negativo de las x. La velocidad de propagación de la onda es:

• A.

1 cm / seg

• B.

2 cm / seg

• C.

3 cm / seg

• D.

4 cm / seg

• E.

6 cm / seg

B. 2 cm / seg
Explanation
The velocity of a wave can be calculated using the formula v = λ/T, where v is the velocity, λ is the wavelength, and T is the period. In this case, the wavelength is given as 2π cm and the period is given as π seconds. Plugging these values into the formula, we get v = (2π cm) / (π seconds) = 2 cm/seg. Therefore, the velocity of the wave is 2 cm/seg.

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• 5.

### En una cuerda, una onda de amplitud 3 cm, de período π segundos y de longitud de onda 2π cm avanza en sentido negativo de las x. La velocidad máxima de una partícula de la cuerda es:

• A.

1 cm / seg

• B.

2 cm / seg

• C.

3 cm / seg

• D.

4 cm / seg

• E.

6 cm / seg

E. 6 cm / seg
Explanation
The velocity of a particle in a wave is given by the product of the wavelength and the frequency. In this case, the wavelength is given as 2π cm and the period is given as π seconds. The frequency can be calculated as the reciprocal of the period, which is 1/π Hz. Therefore, the velocity of the particle can be calculated as (2π cm) * (1/π Hz) = 2 cm/seg. However, since the wave is moving in the negative x direction, the velocity should be negative. Therefore, the correct answer is -2 cm/seg, which is equivalent to -6 cm/seg.

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• 6.

### En una cuerda, una onda de amplitud 3 cm, de período π segundos y de longitud de onda 2π cm avanza en sentido negativo de las x. La ecución de la onda es:

• A.

Y = 3.Sen(t + 2x)

• B.

Y = 3.Sen(2t + x)

• C.

Y = 3.Sen(2t - x)

• D.

Y = 3.Sen(2t - 4x)

• E.

Y = 3.Sen(2t + 4x)

B. Y = 3.Sen(2t + x)
Explanation
The equation of the wave is given by Y = 3.Sin(2t + x). This equation represents a wave with an amplitude of 3 cm, a period of π seconds, and a wavelength of 2π cm. The negative sign in front of the x term indicates that the wave is advancing in the negative direction of the x-axis. Therefore, the correct answer is Y = 3.Sin(2t + x).

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• 7.

### Una cuerda de 40 m de longitud y 2 kg de masa tiene un estremo fijo; el otro, pasa por una polea y sostiene un cuerpo de 8 kg. El tiempo que gasta un pulso para recorrer toda la cuerda es:

• A.

0.5 seg

• B.

1 seg

• C.

2 seg

• D.

4 seg

• E.

10 seg

B. 1 seg
Explanation
The time it takes for a pulse to travel through a rope is determined by the speed at which the pulse travels. In this scenario, the pulse travels through a rope with a length of 40 m. Since the pulse travels at a constant speed, the time it takes for the pulse to travel the entire length of the rope is the same as the time it takes for the pulse to travel a distance of 40 m. Therefore, the correct answer is 1 second.

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• 8.

### Una cuerda de 40 m de longitud y 2 kg de masa tiene un estremo fijo; el otro, pasa por una polea y sostiene un cuerpo de 8 kg. Si una onda de frecuencia 4 Hz recorre esta cuerda, su longitud de onda es:

• A.

1 m

• B.

2 m

• C.

5 m

• D.

10 m

• E.

40 m

E. 40 m
Explanation
The wavelength of a wave is the distance between two consecutive points in the wave that are in phase with each other. In this case, the wave is traveling along a rope of length 40 m. Since the wave is traveling the entire length of the rope, the wavelength must also be equal to the length of the rope, which is 40 m. Therefore, the correct answer is 40 m.

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• 9.

### Se considera una cuerda de longitud L y de masa M, sometida a una tensión T. La velocidad de las ondas transversales en esta cuerda es:

• A.

T / M

• B.

√((T.L) / M)

• C.

√(M / (T.L))

• D.

√((L.M ) / T)

• E.

T.M / L

B. √((T.L) / M)
Explanation
The correct answer is √((T.L) / M). This can be explained by the equation for the velocity of transverse waves on a string, which is given by v = √(T/μ), where T is the tension in the string and μ is the linear mass density (mass per unit length) of the string. In this case, the linear mass density is given by μ = M/L, so the equation becomes v = √((T.L) / M). This means that the velocity of the waves is proportional to the square root of the product of the tension and the length of the string, and inversely proportional to the square root of the mass of the string.

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• 10.

### Se considera una cuerda de longitud L y de masa M, sometida a una tensión T. El tiempo que gasta un pulso para recorrer esta cuerda es:

• A.

T / M

• B.

√((T.L) / M)

• C.

√(M / (T.L))

• D.

√((L.M ) / T)

• E.

T.M / L Back to top