Answer to Question #122357 in Mechanics | Relativity for Lilly

Question

A wave train moves can travel through 76 cm of a tight spring in 0.23 seconds, and it forms
3 crests in a distance of 24.0 cm. It then enters a loose spring and its speed decreases to
1.1 m/s. What is the wavelength in the loose spring? (Which property remains the same?)

Answer 1

Explanations & Calculations

• Velocity of the wave train within the first medium can be calculated to be,

$\qquad\qquad \begin{aligned} \small V_1 &= \small \frac{0.76 m }{0.23s} = \bold{3.30ms^{-1}} \end{aligned}$

• Magnitude of the wavelength is defined as the distance between 2 identical points — one crest to another closest one, one trough to another closest one.
• Therefore, 3 crests in a given distance means 2 wavelengths are generated within.

• Therefore, the wavelength for the first medium ($\lambda_1$ ) could be calculated,

$\qquad\qquad \begin{aligned} ‍\small 2\lambda_1 &= \small 0.24m\\ \small \lambda_1 &= \small 0.12 m \end{aligned}$

• For a wave propagates between several media, the frequency ($f$ ) remains unchanged, only the wavelength changes due to the friction from the medium resulting a change in its velocity.

• Therefore, for the wave within first medium,

$\qquad\qquad \begin{aligned} \small f &= \small \frac{3.3ms^{-1}}{0.12m} \cdots(\because V =f\lambda) \end{aligned}$

• For the second situation,

$\qquad\qquad \begin{aligned} \small f &= \small \frac{1.1ms^{-1}}{\lambda_2} \end{aligned}$

• Since the $f$ is same,

$\qquad\qquad \begin{aligned} \small \frac{3.3ms^{-1}}{0.12m} &= \small \frac{1.1ms^{-1}}{\lambda_2}\\ \small \lambda_2 &= \small \bold{0.04m=4cm} \end{aligned}$