A progressive transverse mechanical wave travels along a stretched string with a wavelengt… — JAMB UTME

A progressive transverse mechanical wave travels along a stretched string with a wavelength of $\lambda = 0.80\text{ m}$ . If the maximum speed of an oscillating particle on the string is equal to one-quarter ( $\frac{1}{4}$ ) of the propagation speed of the wave, what is the amplitude of the wave, and how are the particle vibration and energy propagation directions oriented relative to each other?

The amplitude is $\frac{1}{10\pi}\text{ m}$ , and particles vibrate perpendicular to the direction of energy propagation.Cevap
B
The amplitude is $\frac{1}{5\pi}\text{ m}$ , and particles vibrate perpendicular to the direction of energy propagation.
C
The amplitude is $0.20\text{ m}$ , and particles vibrate parallel to the direction of energy propagation.
D
The amplitude is $\frac{1}{10\pi}\text{ m}$ , and particles vibrate parallel to the direction of energy propagation.

Cevap

The amplitude of the wave is

\frac{1}{10\pi}\text{ m}

, and particles vibrate perpendicular to the direction of energy propagation.

The maximum speed of a particle in simple harmonic wave motion is

v_{p,\text{max}} = \omega A = 2\pi f A

. The wave propagation speed is

v = f \lambda

. Given

v_{p,\text{max}} = \frac{1}{4}v

, we set

2\pi f A = \frac{1}{4} f \lambda

, which yields

A = \frac{\lambda}{8\pi} = \frac{0.80}{8\pi} = \frac{1}{10\pi}\text{ m}

. Because the wave is transverse, particle vibrations occur perpendicular to the direction of wave energy propagation.

Adım Adım Çözüm

Relate maximum particle speed to wave parameters.

Maximum transverse particle speed

v_{p,\text{max}} = \omega A = 2\pi f A

Particles in simple harmonic wave motion have maximum speed given by the product of angular frequency

\omega

and amplitude

A

Express wave propagation speed in terms of frequency and wavelength.

Wave speed

v = f \lambda

The fundamental wave equation relates wave speed

v

directly to frequency

f

and wavelength

\lambda

Set up the given proportion and solve for amplitude

A

2\pi f A = \frac{1}{4} (f \lambda) \implies 2\pi A = \frac{\lambda}{4} \implies A = \frac{\lambda}{8\pi} = \frac{0.80}{8\pi} = \frac{1}{10\pi}\text{ m}

Canceling frequency

f

from both sides allows direct evaluation of

A

Classify the direction of particle motion relative to energy propagation for a transverse wave.

Particles vibrate perpendicular to the direction of wave travel.

By definition, transverse mechanical waves involve oscillations perpendicular to the direction of wave energy propagation.

Anahtar Kavram

Relationship between particle velocity and wave velocity in transverse mechanical waves

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