Concept explainers
Review. A steel guitar string with a diameter of 1.00 mm is stretched between supports 80.0 cm apart. The temperature is 0.0°C. (a) Find the mass per unit length of this siring. (Use the value 7.86 × 103 kg/m4 for the density.) (b) The fundamental frequency of transverse oscillations of the string is 200 Hz. What is the tension in the string? Next, the temperature is raised to 30.0°C. Find the resulting values of (c) the tension and (d) the fundamental frequency. Assume both the Young’s modulus of 20.0 × 1010 N/m2 and the average coefficient of expansion α = 11.0 × 10-6 (°C)-1 have constant values between 0.0°C and 30.0°C.
(a)
The mass per unit length of the string.
Answer to Problem 19.73CP
The mass per unit length of the string is
Explanation of Solution
Given Info: The diameter of the string of the steel guitar is
Formula to calculate the radius of the wire is,
Here,
Substitute
Thus, the value of the radius is
The area of cross section of the steel string is,
Substitute
Thus, the area of cross section of the steel string is
Formula to calculate the mass per unit length of the steel string is,
Here,
Substitute
Conclusion:
Therefore, the mass per unit length of the string is
(b)
The Tension in the string.
Answer to Problem 19.73CP
The Tension in the string is
Explanation of Solution
Given Info: The diameter of the string of the steel guitar is
Formula to calculate the fundamental frequency is,
Here,
Rearrange the above expression for
Substitute
Conclusion:
Therefore, the Tension in the string is
(c)
The Tension in the string when the temperature is raised to
Answer to Problem 19.73CP
The Tension in the string when the temperature is raised to
is
Explanation of Solution
Given Info: The diameter of the string of the steel guitar is
Formula for the change in the length, when temperature varies is,
Here,
Substitute
Thus, the final length of the brass pendulum is
Formula to calculate the tension in the wire is,
Substitute
Conclusion:
Therefore, the Tension in the string when the temperature is raised to
is
(d)
The fundamental frequency of the string.
Answer to Problem 19.73CP
The fundamental frequency of the string is
Explanation of Solution
Given Info: The diameter of the string of the steel guitar is
Write the expression for the fundamental frequency.
Substitute
Conclusion:
Therefore, the fundamental frequency of the string is
Want to see more full solutions like this?
Chapter 19 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
- No chatgpt pls will upvotearrow_forwardair is pushed steadily though a forced air pipe at a steady speed of 4.0 m/s. the pipe measures 56 cm by 22 cm. how fast will air move though a narrower portion of the pipe that is also rectangular and measures 32 cm by 22 cmarrow_forwardNo chatgpt pls will upvotearrow_forward
- 13.87 ... Interplanetary Navigation. The most efficient way to send a spacecraft from the earth to another planet is by using a Hohmann transfer orbit (Fig. P13.87). If the orbits of the departure and destination planets are circular, the Hohmann transfer orbit is an elliptical orbit whose perihelion and aphelion are tangent to the orbits of the two planets. The rockets are fired briefly at the depar- ture planet to put the spacecraft into the transfer orbit; the spacecraft then coasts until it reaches the destination planet. The rockets are then fired again to put the spacecraft into the same orbit about the sun as the destination planet. (a) For a flight from earth to Mars, in what direction must the rockets be fired at the earth and at Mars: in the direction of motion, or opposite the direction of motion? What about for a flight from Mars to the earth? (b) How long does a one- way trip from the the earth to Mars take, between the firings of the rockets? (c) To reach Mars from the…arrow_forwardNo chatgpt pls will upvotearrow_forwarda cubic foot of argon at 20 degrees celsius is isentropically compressed from 1 atm to 425 KPa. What is the new temperature and density?arrow_forward
- Calculate the variance of the calculated accelerations. The free fall height was 1753 mm. The measured release and catch times were: 222.22 800.00 61.11 641.67 0.00 588.89 11.11 588.89 8.33 588.89 11.11 588.89 5.56 586.11 2.78 583.33 Give in the answer window the calculated repeated experiment variance in m/s2.arrow_forwardNo chatgpt pls will upvotearrow_forwardCan you help me solve the questions pleasearrow_forward
- Can you help me solve these questions please so i can see how to do itarrow_forwardHow can i solve this if n1 (refractive index of gas) and n2 (refractive index of plastic) is not known. And the brewsters angle isn't knownarrow_forward2. Consider the situation described in problem 1 where light emerges horizontally from ground level. Take k = 0.0020 m' and no = 1.0001 and find at which horizontal distance, x, the ray reaches a height of y = 1.5 m.arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning