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7 days ago

10

You and your surfing buddy are waiting to catch a wave a few hundred meters off the beach. The waves are conveniently sinusoidal

, and you notice that when you're on the top of one wave and moving toward your friend, she is exactly halfway between you and the trough of the wave. 1.50 seconds later, your friend is at the top of the wave. You estimate the horizontal distance between you and your friend at 8.00 m. (a) What is the frequency of the waves?

2 answers:

Softa [913]7 days ago

8 0

(a) The frequency of the waves is ¹/₆ Hz ≈ 0.167 Hz

(b) The speed of the waves is 5¹/₃ m/s ≈ 5.33 m/s

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Additional explanation

Let's recall the wave speed and wave intensity formula:

$\large {\boxed {v = \lambda f}}$

f = frequency of the wave (Hz)

v = wave speed (m/s)

λ = wavelength (m)

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$\large {\boxed {I = 2 \pi^2 A^2 f^2 \rho v}}$

I = wave intensity (W/m²)

A = wave amplitude (m)

f = wave frequency (Hz)

ρ = medium density (kg/m³)

v = wave speed (m/s)

Let’s solve the problem now!

$\texttt{ }$

Given:

Time period = t = 1.50 seconds

Distance traveled = d = 8.00 m

Question:

(a) What is the frequency of the waves?

(b) What is the wave speed?

Solution:

Question (a):

$t = \frac{1}{4}T$

$1.50 = \frac{1}{4}T$

$T = 4 \times 1.50$

$T = 6 \texttt{ seconds}$

$\texttt{ }$

$f = \frac{1}{T}$

$f = \frac{1}{6} \texttt{ Hz}$

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Question (b):

$d = \frac{1}{4}\lambda$

$8.00 = \frac{1}{4}\lambda$

$\lambda = 8.00 \times 4$

$\lambda = 32 \texttt{ m}$

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$v = \lambda f$

$v = 32 \times \frac{1}{6}$

$v = 5\frac{1}{3} \texttt{ m/s}$

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Learn more

Doppler Effect:
Doppler Effect Example:
Sound Waves Can't Travel In Space:
Beat Frequency - Doppler Effect:

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Answer details

Grade: College

Subject: Physics

Chapter: Sound Waves

Softa [913]7 days ago

6 0

Answer:

(a): The calculated wave frequency is f= 0.16 Hz

Explanation:

T/4= 1.5 s

T= 6 sec

f= 1/T

f= 0.16 Hz (a)

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