Please help with both incorrect parts, show work i am trying to use this to study. **Transverse Wave on a Cord**A transverse wave on a cord is described by the equation:\[ D(x, t) = 0.10 \sin(4.0x - 40.0t) \]where $ D $ and $ x $ are in meters (m) and $ t $ is in seconds (s).---**Part A****Problem:** At $ t = 7.5 \times 10^{-2} \, \text{s} $, what is the displacement of the point on the cord where $ x = 0.45 \, \text{m} $?**Instructions:** Express your answer using two significant figures.**Solution:**\[ D(0.45 \, \text{m}, 7.5 \times 10^{-2} \, \text{s}) = -9.3 \times 10^{-2} \, \text{m} \]**Feedback:** Correct---**Part B****Problem:** At $ t = 7.5 \times 10^{-2} \, \text{s} $, what is the velocity of the point on the cord where $ x = 0.45 \, \text{m} $?**Instructions:** Express your answer using two significant figures.**Solution:**\[\frac{\partial D}{\partial t} (0.45 \, \text{m}, 7.5 \times 10^{-2} \, \text{s}) = -7.5 \times 10^{-2} \, \text{m/s} \]**Feedback:** Incorrect; Try Again: 3 attempts remaining---This section aims to help students understand and calculate the displacement and velocity of a point on a cord experiencing a transverse wave. ### Understanding Beat Frequency in PipesIn this example, we examine two identical tubes, each closed at one end, with a fundamental frequency of 350 Hz at 21.0°C. The air temperature for one of the tubes is increased to 32.0°C. **Part A: Calculating Beat Frequency**When the two pipes are sounded together, we need to determine the resulting beat frequency.**Instructions:** Express your answer using two significant figures.**Given Answer:** \[ f_{\text{beat}} = 7.5 \, \text{Hz} \]**Feedback:**The given answer is incorrect. Please try again; you have 2 attempts remaining.

Answered: Part B At t = 7.5 x 10-²s, what is the…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Question

Please help with both incorrect parts, show work i am trying to use this to study.

**Transverse Wave on a Cord**

A transverse wave on a cord is described by the equation:

\[ D(x, t) = 0.10 \sin(4.0x - 40.0t) \]

where $ D $ and $ x $ are in meters (m) and $ t $ is in seconds (s).

---

**Part A**

**Problem:** At $ t = 7.5 \times 10^{-2} \, \text{s} $, what is the displacement of the point on the cord where $ x = 0.45 \, \text{m} $?

**Instructions:** Express your answer using two significant figures.

**Solution:**

\[ D(0.45 \, \text{m}, 7.5 \times 10^{-2} \, \text{s}) = -9.3 \times 10^{-2} \, \text{m} \]

**Feedback:** Correct

---

**Part B**

**Problem:** At $ t = 7.5 \times 10^{-2} \, \text{s} $, what is the velocity of the point on the cord where $ x = 0.45 \, \text{m} $?

**Instructions:** Express your answer using two significant figures.

**Solution:**

\[\frac{\partial D}{\partial t} (0.45 \, \text{m}, 7.5 \times 10^{-2} \, \text{s}) = -7.5 \times 10^{-2} \, \text{m/s} \]

**Feedback:** Incorrect; Try Again: 3 attempts remaining

---

This section aims to help students understand and calculate the displacement and velocity of a point on a cord experiencing a transverse wave.

### Understanding Beat Frequency in Pipes

In this example, we examine two identical tubes, each closed at one end, with a fundamental frequency of 350 Hz at 21.0°C. The air temperature for one of the tubes is increased to 32.0°C.

**Part A: Calculating Beat Frequency**

When the two pipes are sounded together, we need to determine the resulting beat frequency.

**Instructions:**
Express your answer using two significant figures.

**Given Answer:**
\[ f_{\text{beat}} = 7.5 \, \text{Hz} \]

**Feedback:**
The given answer is incorrect. Please try again; you have 2 attempts remaining.

Expert Solution

Step 1

Given,

Displacement equation

Step 2

1.Part (B)

The velocity equation is obtained by

$v (x, t) = \frac{d (D (x, t))}{d t} v (x, t) = \frac{d (0.10 \sin (4.0 x - 40.0 t))}{d t} v (x, t) = 0.10 \cos (4.0 x - 40.0 t) * (- 40.0) v (x, t) = - 4.0 \cos (4.0 x - 40.0 t) v (x = 0.45 m, t = 7.5 * 10^{- 2} s) = - 4.0 \cos (4.0 * 0.45 - 40.0 * 7.5 * 10^{- 2}) v (x = 0.45 m, t = 7.5 * 10^{- 2} s) = - 4.0 \cos (1.8 - 3) v (x = 0.45 m, t = 7.5 * 10^{- 2} s) = - 4 \cos (- 1.2) v (x = 0.45 m, t = 7.5 * 10^{- 2} s) = - 1.45 m / s v (x = 0.45 m, t = 7.5 * 10^{- 2} s) = - 1.5 m / s$

Trending now

This is a popular solution!

Step by step

Solved in 4 steps

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions