November 15, 2019

Solution Set 4

Physics 208-02 – Introductory Physics II
Solution Set 4
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
www.CallMeDrRob.com
Spring 2012

§ Chapter 11, Problem 23

At time t=0 a 755\;\mathrm{g} mass at rest on the end of a horizontal spring (k=124\;\mathrm{N}/\mathrm{m}) is struck by a hammer which gives the mass an initial speed of 2.96\;\mathrm{m}/\mathrm{s}.

§ (a)

Determine the period and frequency of the motion.

\displaystyle\omega \displaystyle= \displaystyle\sqrt{\frac{k}{m}}
\displaystyle= \displaystyle\sqrt{\frac{124\frac{\mathrm{N}}{\mathrm{m}}}{0.755\;\mathrm{kg}}}
\displaystyle= \displaystyle 12.\underline{8}2\;\mathrm{Hz}
\displaystyle f \displaystyle= \displaystyle\frac{\omega}{2\pi}
\displaystyle= \displaystyle\frac{12.\underline{8}2\;\mathrm{Hz}}{2\pi}
\displaystyle= \displaystyle 2.0\underline{4}0\;\mathrm{Hz}
\displaystyle T \displaystyle= \displaystyle\frac{1}{f}
\displaystyle= \displaystyle\frac{1}{2.0\underline{4}0\;\mathrm{Hz}}
\displaystyle= \displaystyle 0.490\;\mathrm{s}

§ (b)

Determine the amplitude of the motion.

Initially, immediately after being struck, it is at the equilibrium position, which means the positions of maximum velocity.

\displaystyle v_{\mathrm{max}} \displaystyle= \displaystyle\omega A
\displaystyle A \displaystyle= \displaystyle\frac{v_{\mathrm{max}}}{\omega}
\displaystyle= \displaystyle\frac{2.96\frac{\mathrm{m}}{\mathrm{s}}}{12.\underline{8}2\;\mathrm{Hz}}
\displaystyle= \displaystyle 0.231\;\mathrm{m}

§ (c)

Determine the maximum acceleration.

Combine Hooke’s Law, F=-kx, with Newton’s Second Law, F=ma.

\displaystyle ma \displaystyle= \displaystyle-kx

We’ll get the largest acceleration when the displacement is a maximum. Taking just the magnitude,

\displaystyle a_{\mathrm{max}} \displaystyle= \displaystyle\frac{k}{m}A
\displaystyle= \displaystyle\omega^{2}A
\displaystyle= \displaystyle\left(12.\underline{8}2\;\mathrm{Hz}\right)^{2}\left(0.231\;\mathrm{m}\right)
\displaystyle= \displaystyle 37.9\frac{\mathrm{m}}{\mathrm{s}^{2}}

§ (d)

Determine the total energy.

We can compute this using v_{\mathrm{max}} when all of the energy is kinetic, or we can use A when all of the energy is potential.

\displaystyle E \displaystyle= \displaystyle\frac{1}{2}mv_{\mathrm{max}}^{2}
\displaystyle= \displaystyle\frac{1}{2}(0.755\;\mathrm{kg})\left(2.96\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}
\displaystyle= \displaystyle 3.31\;\mathrm{J}
\displaystyle E \displaystyle= \displaystyle\frac{1}{2}kA^{2}
\displaystyle= \displaystyle\frac{1}{2}\left(124\frac{\mathrm{N}}{\mathrm{m}}\right)\left(0.231\;\mathrm{m}\right)^{2}
\displaystyle= \displaystyle 3.31\;\mathrm{J}

§ Chapter 11, Problem 52

If a violin string vibrates at 440\;\mathrm{Hz} as its fundamental frequency, what are its first four harmonics?

Use

\displaystyle f_{n} \displaystyle= \displaystyle n\frac{v}{2L}
\displaystyle= \displaystyle nf_{1}

Then

\displaystyle f_{1} \displaystyle= \displaystyle 440\;\mathrm{Hz}
\displaystyle f_{2} \displaystyle= \displaystyle 2f_{1}
\displaystyle= \displaystyle 2\left(440\;\mathrm{Hz}\right)
\displaystyle= \displaystyle 880\;\mathrm{Hz}
\displaystyle f_{3} \displaystyle= \displaystyle 3f_{1}
\displaystyle= \displaystyle 3\left(440\;\mathrm{Hz}\right)
\displaystyle= \displaystyle 1320\;\mathrm{Hz}
\displaystyle f_{4} \displaystyle= \displaystyle 4f_{1}
\displaystyle= \displaystyle 4\left(440\;\mathrm{Hz}\right)
\displaystyle= \displaystyle 1760\;\mathrm{Hz}

§ Chapter 11, Problem 53

A violin string vibrates at 294\;\mathrm{Hz} when unfingered. At what frequency will it vibrate when fingered one-third of the way down the from the end such that the vibrating length is two-thirds of the original length?

\displaystyle f_{1} \displaystyle= \displaystyle\frac{v}{2L}
\displaystyle f_{2} \displaystyle= \displaystyle\frac{v}{2\left(\frac{2}{3}L\right)}
\displaystyle\frac{f_{1}}{f_{2}} \displaystyle= \displaystyle\frac{\frac{v}{2L}}{\frac{v}{2\left(\frac{2}{3}L\right)}}
\displaystyle= \displaystyle\frac{v}{2L}\frac{2\left(\frac{2}{3}L\right)}{v}
\displaystyle= \displaystyle\frac{2}{3}
\displaystyle f_{2} \displaystyle= \displaystyle\frac{3}{2}f_{1}

§ Chapter 11, Problem 55

The velocity of waves on a string is 92\;\mathrm{m}/\mathrm{s}. If the frequency of standing waves is 475\;\mathrm{Hz}, how far apart are the adjacent nodes?

\displaystyle\lambda f \displaystyle= \displaystyle v
\displaystyle\lambda \displaystyle= \displaystyle\frac{v}{f}
\displaystyle= \displaystyle\frac{92\frac{\mathrm{m}}{\mathrm{s}}}{475\;\mathrm{Hz}}
\displaystyle= \displaystyle 0.194\;\mathrm{m}

The distance between nodes is one-half wavelength, or 9.7\;\mathrm{cm}.

§ Chapter 11, Problem 56

If two successive overtones of a violin are 280\;\mathrm{Hz} and 350\;\mathrm{Hz}, what is the frequency of the fundamental?

Use

\displaystyle f_{n} \displaystyle= \displaystyle n\frac{v}{2L}
\displaystyle= \displaystyle nf_{1}
\displaystyle f_{{n+1}}-f_{n} \displaystyle= \displaystyle(n+1)f_{n}-nf_{n}
\displaystyle= \displaystyle f_{n}

So

\displaystyle f_{1} \displaystyle= \displaystyle 350\;\mathrm{Hz}-280\;\mathrm{Hz}
\displaystyle= \displaystyle 70\;\mathrm{Hz}

§ Chapter 11, Problem 57

A guitar string is 90\;\mathrm{cm} long and has a mass of 3.6\;\mathrm{g}. The distance from the bridge to the support post is L=62\;\mathrm{cm}, and the string is under a tension of 520\;\mathrm{N}. What are the frequencies of the fundamental and first two overtones?

\displaystyle v \displaystyle= \displaystyle\sqrt{\frac{T}{m/l}}
\displaystyle= \displaystyle\sqrt{\frac{520\;\mathrm{N}}{0.0036\;\mathrm{kg}/0.90\;\mathrm{m}}}
\displaystyle= \displaystyle 360\frac{\mathrm{m}}{\mathrm{s}}
\displaystyle f_{1} \displaystyle= \displaystyle\frac{v}{2L^{\prime}}
\displaystyle= \displaystyle\frac{360\frac{\mathrm{m}}{\mathrm{s}}}{2(0.62\;\mathrm{m})}
\displaystyle= \displaystyle 290\;\mathrm{Hz}
\displaystyle f_{2} \displaystyle= \displaystyle 2f_{1}
\displaystyle= \displaystyle 2\left(290\;\mathrm{Hz}\right)
\displaystyle= \displaystyle 580\;\mathrm{Hz}
\displaystyle f_{3} \displaystyle= \displaystyle 3f_{1}
\displaystyle= \displaystyle 3\left(290\;\mathrm{Hz}\right)
\displaystyle= \displaystyle 870\;\mathrm{Hz}

§ Chapter 11, Problem 58

A particular guitar string is supposed to vibrate at 200\;\mathrm{Hz}, but it is measured to vibrate at 205\;\mathrm{Hz}. By what percent should the tension in the string be changed to correct the frequency?

\displaystyle v \displaystyle= \displaystyle\sqrt{\frac{T}{m/L}}
\displaystyle\lambda f \displaystyle= \displaystyle v
\displaystyle= \displaystyle\sqrt{\frac{T}{m/L}}

Write this equation twice realizing that the wavelength and the mass per length do not change.

\displaystyle\frac{\lambda f_{1}}{\lambda f_{2}} \displaystyle= \displaystyle\frac{\sqrt{\frac{T_{1}}{m/L}}}{\sqrt{\frac{T_{2}}{m/L}}}
\displaystyle\frac{f_{1}}{f_{2}} \displaystyle= \displaystyle\sqrt{\frac{T_{1}}{T_{2}}}
\displaystyle\left(\frac{f_{1}}{f_{2}}\right)^{2} \displaystyle= \displaystyle\frac{T_{1}}{T_{2}}

To compute the percentage change,

\displaystyle\frac{T_{2}-T_{1}}{T_{1}}\times 100\;\mathrm{percent} \displaystyle= \displaystyle\left(\frac{T_{2}}{T_{1}}-1\right)\times 100\;\mathrm{percent}
\displaystyle= \displaystyle\left(\left(\frac{f_{2}}{f_{1}}\right)^{2}-1\right)\times 100\;\mathrm{percent}
\displaystyle= \displaystyle\left(\left(\frac{200\;\mathrm{Hz}}{205\;\mathrm{Hz}}\right)^{2}-1\right)\times 100\;\mathrm{percent}
\displaystyle= \displaystyle-5\;\mathrm{percent}

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