November 15, 2019

Exam 2

Physics 208 – Introductory Physics II
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
Spring 2012

Exam 2
Oscillations & Waves

§ Problem 1

The motion of a 500\;\mathrm{g} mass hanging on a spring with spring constant k is given by the following equation:

\displaystyle z=1.10\;\mathrm{m}\cos\left(4.19\;\mathrm{s}^{{-1}}\, t\right)

§ (a)

What is the amplitude, A, of the motion, and what is the total distance traveled during the first cycle? (A simple sketch may be useful.)
(5 points)

Compare with the form

\displaystyle z=A\cos\left(\omega t\right)


\displaystyle A=1.10\;\mathrm{m}

In one full cycle it travels from z=A through z=0 to z=-A, then back through z=0 to z=A. Hence the total distance traveled is 4A.

\displaystyle\mathrm{distance} \displaystyle= \displaystyle 4A
\displaystyle= \displaystyle 4(1.10\;\mathrm{m})
\displaystyle= \displaystyle 4.40\;\mathrm{m}

§ (b)

What is the frequency, f, of the oscillations?
(5 points)

Compare with the form

\displaystyle z=A\cos\left(\omega t\right)


\displaystyle\omega \displaystyle= \displaystyle 4.19\;\mathrm{s}^{{-1}}
\displaystyle f \displaystyle= \displaystyle\frac{\omega}{2\pi}
\displaystyle= \displaystyle\frac{4.19\;\mathrm{s}^{{-1}}}{2\pi}
\displaystyle= \displaystyle 0.66\underline{6}8\;\mathrm{Hz}
\displaystyle= \displaystyle 0.667\;\mathrm{Hz}

§ (c)

What is the value of the spring constant, k?
(5 points)

\displaystyle\omega \displaystyle= \displaystyle\sqrt{\frac{k}{m}}
\displaystyle k \displaystyle= \displaystyle m\omega^{2}
\displaystyle= \displaystyle(0.5\;\mathrm{kg})\;\left(4.19\;\mathrm{s}^{{-1}}\right)^{2}
\displaystyle= \displaystyle\underline{8}.8\;\frac{\mathrm{N}}{\mathrm{m}}
\displaystyle= \displaystyle 9\;\frac{\mathrm{N}}{\mathrm{m}}

§ (d)

What is the maximum speed, v_{\mathrm{max}}, of the mass?
(5 points)

\displaystyle v_{\mathrm{max}} \displaystyle= \displaystyle\omega A
\displaystyle= \displaystyle\left(4.19\;\mathrm{s}^{{-1}}\right)(1.10\;\mathrm{m})
\displaystyle= \displaystyle 4.\underline{6}1\;\frac{\mathrm{m}}{\mathrm{s}}
\displaystyle= \displaystyle 4.6\;\frac{\mathrm{m}}{\mathrm{s}}

§ (e)

What is the maximum acceleration, a_{\mathrm{max}}, of the mass?
(5 points)

\displaystyle a_{\mathrm{max}} \displaystyle= \displaystyle\omega^{2}A
\displaystyle= \displaystyle\left(4.19\;\mathrm{s}^{{-1}}\right)^{2}(1.10\;\mathrm{m})
\displaystyle= \displaystyle 1\underline{9}.3\;\frac{\mathrm{m}}{\mathrm{s}^{2}}
\displaystyle= \displaystyle 19\;\frac{\mathrm{m}}{\mathrm{s}^{2}}

§ (f)

At what time does the mass first reach the position z=0.9\;\mathrm{m}?
(5 points)

\displaystyle 0.9\;\mathrm{m} \displaystyle= \displaystyle 1.10\;\mathrm{m}\cos\left(4.19\;\mathrm{s}^{{-1}}\, t\right)
\displaystyle\cos\left(4.19\;\mathrm{s}^{{-1}}\, t\right) \displaystyle= \displaystyle 0.\underline{8}2
\displaystyle 4.19\;\mathrm{s}^{{-1}}\, t \displaystyle= \displaystyle\cos^{{-1}}(0.\underline{8}2)
\displaystyle= \displaystyle 0.\underline{6}1
\displaystyle t \displaystyle= \displaystyle\frac{0.\underline{6}1}{4.19\;\mathrm{s}^{{-1}}}
\displaystyle= \displaystyle 0.\underline{1}46\;\mathrm{s}

WARNING: Your calculator should be in radian mode! Otherwise you need to convert from degrees to radians for a physically meaningful result.

§ (g)

What is the position of the mass at t=30\;\mathrm{s}?
(5 points)

\displaystyle z \displaystyle= \displaystyle 1.10\;\mathrm{m}\cos\left(4.19\;\mathrm{s}^{{-1}}\,(30\;\mathrm{s})\right)
\displaystyle= \displaystyle\underline{1}.10\;\mathrm{m}

WARNING: Your calculator should be in radian mode!

There is another way to see this solution. Above we calculated the frequency, f. From this we can get the period, T.

\displaystyle T \displaystyle= \displaystyle\frac{1}{f}
\displaystyle= \displaystyle\frac{1}{0.66\underline{6}8\;\mathrm{Hz}}
\displaystyle= \displaystyle 1.50\;\mathrm{s}

If we observe that

\displaystyle\frac{30\;\mathrm{s}}{T} \displaystyle= \displaystyle\frac{30\;\mathrm{s}}{1.50\;\mathrm{s}}
\displaystyle= \displaystyle 20

we see that the mass has undergone 20 complete cycles, hence it should be exactly where it was at time t=0\;\mathrm{s}.

\displaystyle z \displaystyle= \displaystyle\underline{1}.10\;\mathrm{m}

§ Problem 2

Working with a guitar string in the lab a student finds a harmonic with many nodes at 2640\;\mathrm{Hz}. The next harmonic occurs at 3080\;\mathrm{Hz}. The string has a total length of 1.2\;\mathrm{m}, but the portion that vibrates between the first end to where it passes over a pulley is 1.0\;\mathrm{m}. The tension is created by hanging a 5.0\;\mathrm{kg} mass from the second end.

§ (a)

What is the fundamental frequency for this string? (10 points)

Recall that

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

If we consider two adjacent harmonics, then

\displaystyle f_{{n+1}}-f_{n} \displaystyle= \displaystyle(n+1)f_{1}-nf_{1}
\displaystyle= \displaystyle f_{1}

All we must do to find f_{1} is to difference two adjacent frequencies.

\displaystyle f_{1} \displaystyle= \displaystyle 3080\;\mathrm{Hz}-2640\;\mathrm{Hz}
\displaystyle= \displaystyle 440\;\mathrm{Hz}

§ (b)

Which two harmonics were observed by the student in the lab? (10 points)

Recall that

\displaystyle f_{n} \displaystyle= \displaystyle nf_{1}
\displaystyle n \displaystyle= \displaystyle\frac{f_{n}}{f_{1}}
\displaystyle= \displaystyle\frac{2640\;\mathrm{Hz}}{440\;\mathrm{Hz}}
\displaystyle= \displaystyle 6
\displaystyle n+1 \displaystyle= \displaystyle 7

The student observed the 6^{\mathrm{th}} and 7^{\mathrm{th}} harmonics.

§ (c)

What is the mass m of the wire? (10 points)

First let’s find the wave speed using

\displaystyle v \displaystyle= \displaystyle\lambda f

noting that the wavelength for the fundamental is twice the length of the vibrating portion of the string. So

\displaystyle v \displaystyle= \displaystyle 2\left(1.0\;\mathrm{m}\right)\left(440\;\mathrm{s}^{{-1}}\right)
\displaystyle= \displaystyle 880\frac{\mathrm{m}}{\mathrm{s}}

Next we note that the wave velocity may be determined from the string length, mass and tension as

\displaystyle v \displaystyle= \displaystyle\sqrt{\frac{T}{m/L}}
\displaystyle v^{2} \displaystyle= \displaystyle\frac{T}{m/L}
\displaystyle m \displaystyle= \displaystyle\frac{TL}{v^{2}}
\displaystyle= \displaystyle\frac{(5.0\;\mathrm{kg})\left(9.81\frac{\mathrm{m}}{\mathrm{s}^{2}}\right)(1.2\;\mathrm{m})}{\left(880\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}}
\displaystyle= \displaystyle 7.6\times 10^{{-5}}\;\mathrm{kg}\times\frac{10^{6}\;\mathrm{mg}}{\mathrm{kg}}
\displaystyle= \displaystyle 76\;\mathrm{mg}

§ Problem 3

As shown in the picture, a block with unknown mass m is attached to a spring with an unknown spring constant k all placed on a frictionless table.

§ (a)

A second 5.0\;\mathrm{kg} block is attached to the first by a string and allowed to hang off the end of the table with the string routed over a frictionless pulley causing the spring to stretch by 5.0\;\mathrm{cm}. What is the spring constant k? (10 points)

The Hooke’s Law spring force will be equal to the weight of the suspended block. (Transmitted through the string tension.)

\displaystyle kx \displaystyle= \displaystyle mg
\displaystyle k \displaystyle= \displaystyle\frac{mg}{x}
\displaystyle= \displaystyle\frac{(5.0\;\mathrm{kg})\left(9.81\frac{\mathrm{m}}{\mathrm{s}^{2}}\right)}{0.050\;\mathrm{m}}
\displaystyle= \displaystyle 9\underline{8}1\frac{\mathrm{N}}{\mathrm{m}}
\displaystyle= \displaystyle 980\frac{\mathrm{N}}{\mathrm{m}}

§ (b)

The string is cut, and the block begins to oscillate with a frequency of 0.55\;\mathrm{s}^{{-1}}. What is the mass of the block? (5 points)

\displaystyle\omega \displaystyle= \displaystyle\sqrt{\frac{k}{m}}
\displaystyle(2\pi f) \displaystyle= \displaystyle\sqrt{\frac{k}{m}}
\displaystyle m \displaystyle= \displaystyle\frac{k}{(2\pi f)^{2}}
\displaystyle= \displaystyle\frac{9\underline{8}1\frac{\mathrm{N}}{\mathrm{m}}}{4\pi^{2}\left(0.55\;\mathrm{s}^{{-1}}\right)^{2}}
\displaystyle= \displaystyle 82\;\mathrm{kg}

§ (c)

What is the maximum speed, v_{\mathrm{max}}, obtained by the block? (10 points)

\displaystyle v_{\mathrm{max}} \displaystyle= \displaystyle\omega A
\displaystyle= \displaystyle 2\pi fA
\displaystyle= \displaystyle 2\pi\left(0.55\;\mathrm{s}^{{-1}}\right)\left(0.050\;\mathrm{m}\right)
\displaystyle= \displaystyle 0.17\frac{\mathrm{m}}{\mathrm{s}}

§ Problem 4

Planet g
Mercury 3.61\frac{\mathrm{m}}{\mathrm{s}^{2}}
Venus 8.83\frac{\mathrm{m}}{\mathrm{s}^{2}}
Earth 9.81\frac{\mathrm{m}}{\mathrm{s}^{2}}
Mars 3.75\frac{\mathrm{m}}{\mathrm{s}^{2}}
Jupiter 26.0\frac{\mathrm{m}}{\mathrm{s}^{2}}
Saturn 11.2\frac{\mathrm{m}}{\mathrm{s}^{2}}
Uranus 10.5\frac{\mathrm{m}}{\mathrm{s}^{2}}
Neptune 13.3\frac{\mathrm{m}}{\mathrm{s}^{2}}
Pluto 0.61\frac{\mathrm{m}}{\mathrm{s}^{2}}

Table 1.

Gravitational acceleration on bodies in our solar system.

A 20.0\;\mathrm{cm} pendulum undergoes approximately 17 complete oscillations in one full minute. According to Table 1, where in the solar system is the pendulum most likely located? (No credit for guessing without a sensible calculation.)

We know that

\displaystyle\omega \displaystyle= \displaystyle\frac{g}{L}
\displaystyle g \displaystyle= \displaystyle L\omega^{2}
\displaystyle= \displaystyle L\left(2\pi f\right)^{2}
\displaystyle= \displaystyle L\left(\frac{2\pi}{T}\right)^{2}

We may obtain the period from the information given.

\displaystyle T \displaystyle= \displaystyle\frac{60\;\mathrm{s}}{17}
\displaystyle= \displaystyle\underline{3}.53\;\mathrm{s}

and hence

\displaystyle\omega \displaystyle= \displaystyle\frac{g}{L}
\displaystyle g \displaystyle= \displaystyle(0.200\;\mathrm{m})\left(\frac{2\pi}{\underline{3}.53\;\mathrm{s}}\right)^{2}
\displaystyle= \displaystyle 0.\underline{6}3\frac{\mathrm{m}}{\mathrm{s}^{2}}

The answer, within the one significant figure obtained, is Pluto.


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