November 15, 2019

Solution Set 3

Physics 208-02 – Introductory Physics II
Solution Set 3
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
www.CallMeDrRob.com
Spring 2012

§ Chapter 11, Problem 4

A fisherman’s scale stretches 3.6\;\mathrm{cm} when a 2.7\;\mathrm{kg} fish hangs from it.

§ (a)

What is the spring stiffness constant, k?

The weight of the fish is exactly balanced by the Hooke’s Law force of the spring in the scale.

\displaystyle mg \displaystyle= \displaystyle kx
\displaystyle k \displaystyle= \displaystyle\frac{mg}{x}
\displaystyle= \displaystyle\frac{\left(2.7\;\mathrm{kg}\right)\left(9.8\frac{\mathrm{m}}{\mathrm{s}^{2}}\right)}{0.036\;\mathrm{m}}
\displaystyle= \displaystyle 7\underline{3}5\frac{\mathrm{N}}{\mathrm{m}}
\displaystyle= \displaystyle 740\frac{\mathrm{N}}{\mathrm{m}}

§ (b)

What will be the amplitude and frequency of vibration if the fish is pulled down 2.5\;\mathrm{cm} more and released?

In part (a) the fish is in the equilibrium position. The amplitude is the maximum displacement from equilibrium, hence

\displaystyle A \displaystyle= \displaystyle 2.5\;\mathrm{cm}

The frequency is given by

\displaystyle f \displaystyle= \displaystyle\frac{1}{2\pi}\sqrt{\frac{k}{m}}
\displaystyle= \displaystyle\frac{1}{2\pi}\sqrt{\frac{7\underline{3}5\frac{\mathrm{N}}{\mathrm{m}}}{2.7\;\mathrm{kg}}}
\displaystyle= \displaystyle 2.6\;\mathrm{Hz}

§ Chapter 11, Problem 9

A 0.60\;\mathrm{kg} mass at the end of a spring vibrates 3.0 times per second with an amplitude of 0.13\;\mathrm{m}.

§ (a)

Determine the velocity when it passes through the equilibrium point.

There are multiple ways to approach this problem. If we start with

\displaystyle x \displaystyle= \displaystyle A\cos\omega t
\displaystyle v \displaystyle= \displaystyle v_{\mathrm{max}}\sin\omega t
\displaystyle= \displaystyle\omega A\sin\omega t

First calculate

\displaystyle\omega \displaystyle= \displaystyle 2\pi f
\displaystyle= \displaystyle 2\pi\left(3.0\;\mathrm{Hz}\right)
\displaystyle= \displaystyle 1\underline{8}.8\;\mathrm{Hz}
\displaystyle= \displaystyle 19\;\mathrm{Hz}

Then

\displaystyle v_{\mathrm{max}} \displaystyle= \displaystyle\omega A
\displaystyle= \displaystyle\left(1\underline{8}.8\;\mathrm{Hz}\right)\left(0.13\;\mathrm{m}\right)
\displaystyle= \displaystyle 2.\underline{4}4\frac{\mathrm{m}}{\mathrm{s}}
\displaystyle= \displaystyle 2.4\frac{\mathrm{m}}{\mathrm{s}}

§ (b)

Determine the velocity when it is 0.10\;\mathrm{m} from the equilibrium point.

Use conservation of energy.

\displaystyle E_{\mathrm{total}} \displaystyle= \displaystyle\frac{1}{2}mv^{2}+\frac{1}{2}kx^{2}

Now we may solve for the velocity at the indicated position.

\displaystyle\frac{1}{2}mv^{2} \displaystyle= \displaystyle E_{\mathrm{total}}-\frac{1}{2}kx^{2}
\displaystyle v^{2} \displaystyle= \displaystyle 2\frac{E_{\mathrm{total}}}{m}-\frac{k}{m}x^{2}
\displaystyle= \displaystyle 2\frac{E_{\mathrm{total}}}{m}-\omega^{2}x^{2}
\displaystyle v \displaystyle= \displaystyle\sqrt{2\frac{E_{\mathrm{total}}}{m}-\omega^{2}x^{2}}
\displaystyle= \displaystyle\sqrt{2\frac{1.\underline{7}9\;\mathrm{J}}{0.60\;\mathrm{kg}}-\left(1\underline{8}.8\;\mathrm{Hz}\right)^{2}\left(0.10\;\mathrm{m}\right)^{2}}
\displaystyle= \displaystyle 1.6\;\frac{\mathrm{m}}{\mathrm{s}}

§ (c)

Determine the total energy of the system.

In part (a) we calculated v_{\mathrm{max}} which is the speed when the mass is passing through the equilibrium position and the total energy is equal to the kinetic energy. Hence

\displaystyle E_{\mathrm{total}} \displaystyle= \displaystyle\frac{1}{2}mv_{\mathrm{max}}^{2}
\displaystyle= \displaystyle\frac{1}{2}\left(0.60\;\mathrm{kg}\right)\left(2.\underline{4}4\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}
\displaystyle= \displaystyle 1.\underline{7}9\;\mathrm{J}
\displaystyle= \displaystyle 1.8\;\mathrm{J}

§ (d)

Determine the equation describing the motion of the mass assuming that x was a maximum at t=0.

If x is a maximum at t=0, then we know it is appropriate to use a cosine function which is also a maximum when its argument is zero.

\displaystyle x \displaystyle= \displaystyle A\cos\left(\omega t\right)
\displaystyle= \displaystyle\left(0.13\;\mathrm{m}\right)\cos\left(\frac{1\underline{8}.8}{\mathrm{s}}t\right)

§ Chapter 11, Problem 13

An object with a mass of 3.0\;\mathrm{kg} is attached to a spring with spring stiffness constant k=280\;\mathrm{N}/\mathrm{m} and is executing simple harmonic motion. When the object is 0.020\;\mathrm{m} from its equilibrium position it is moving with a speed of 0.55\;\mathrm{m}/\mathrm{s}.

§ (a)

Calculate the amplitude of the motion.

Use conservation of energy.

\displaystyle E_{\mathrm{total}} \displaystyle= \displaystyle\frac{1}{2}mv^{2}+\frac{1}{2}kx^{2}
\displaystyle= \displaystyle\frac{1}{2}\left(3.0\;\mathrm{kg}\right)\left(0.55\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}+\frac{1}{2}\left(280\frac{\mathrm{N}}{\mathrm{m}}\right)\left(0.020\;\mathrm{m}\right)^{2}
\displaystyle= \displaystyle 0.51\;\mathrm{J}

When the mass is at its full amplitude, then the energy is all potential energy.

\displaystyle\frac{1}{2}kA^{2} \displaystyle= \displaystyle E_{\mathrm{total}}
\displaystyle A \displaystyle= \displaystyle\sqrt{2\frac{E_{\mathrm{total}}}{k}}
\displaystyle= \displaystyle\sqrt{2\frac{0.51\;\mathrm{J}}{280\frac{\mathrm{N}}{\mathrm{m}}}}
\displaystyle= \displaystyle 0.060\;\mathrm{m}

§ (b)

Calculate the maximum velocity attained by the object.

The maximum velocity is achieved when the mass is passing through the equilibrium point and the energy is all in the form of kinetic energy.

\displaystyle\frac{1}{2}mv_{\mathrm{max}}^{2} \displaystyle= \displaystyle E_{\mathrm{total}}
\displaystyle v_{\mathrm{max}} \displaystyle= \displaystyle\sqrt{2\frac{E_{\mathrm{total}}}{m}}
\displaystyle= \displaystyle\sqrt{2\frac{0.51\;\mathrm{J}}{3.0\;\mathrm{kg}}}
\displaystyle= \displaystyle 0.58\frac{\mathrm{m}}{\mathrm{s}}

§ Chapter 11, Problem 16

A 0.60\;\mathrm{kg} mass vibrates according to the equation

\displaystyle x=(0.45\;\mathrm{m})\cos(6.40\;\mathrm{s}^{{-1}}\, t)

which we will compare to the standard form

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

§ (a)

Determine the amplitude.

\displaystyle A=0.45\;\mathrm{m}

§ (b)

Determine the frequency.

\displaystyle\omega \displaystyle= \displaystyle 6.40\;\mathrm{Hz}
\displaystyle f \displaystyle= \displaystyle\frac{\omega}{2\pi}
\displaystyle= \displaystyle\frac{6.40\;\mathrm{Hz}}{2\pi}
\displaystyle= \displaystyle 1.02\;\mathrm{Hz}

§ (c)

Determine the total energy.

\displaystyle E_{\mathrm{total}} \displaystyle= \displaystyle\frac{1}{2}kA^{2}

Since

\displaystyle\omega \displaystyle= \displaystyle\sqrt{\frac{k}{m}}
\displaystyle k \displaystyle= \displaystyle m\omega^{2}

we may write

\displaystyle E_{\mathrm{total}} \displaystyle= \displaystyle\frac{1}{2}m\omega^{2}A^{2}
\displaystyle= \displaystyle\frac{1}{2}\left(0.60\;\mathrm{kg}\right)\left(6.40\;\mathrm{s}^{{-1}}\right)^{2}\left(0.45\;\mathrm{m}\right)^{2}
\displaystyle= \displaystyle 2.49\;\mathrm{J}

§ (d)

Determine the kinetic and potential energies when x=0.30\;\mathrm{m}.

\displaystyle E_{\mathrm{potential}} \displaystyle= \displaystyle\frac{1}{2}kx^{2}
\displaystyle= \displaystyle\frac{1}{2}m\omega^{2}x^{2}
\displaystyle= \displaystyle\frac{1}{2}\left(0.60\;\mathrm{kg}\right)\left(6.40\;\mathrm{s}^{{-1}}\right)^{2}\left(0.30\;\mathrm{m}\right)^{2}
\displaystyle= \displaystyle 1.11\;\mathrm{J}
\displaystyle E_{\mathrm{kinetic}} \displaystyle= \displaystyle E_{\mathrm{total}}-E_{\mathrm{potential}}
\displaystyle= \displaystyle 2.48\;\mathrm{J}-1.11\;\mathrm{J}
\displaystyle= \displaystyle 1.38\;\mathrm{J}

§ Chapter 11, Problem 30

A pendulum has a period of 0.80\;\mathrm{s} on Earth. What is its period on Mars where the acceleration of gravity is 0.37 that on Earth?

This is easiest if we take a ratio.

\displaystyle\frac{T_{\mathrm{Mars}}}{T_{\mathrm{Earth}}} \displaystyle= \displaystyle\frac{\frac{1}{2\pi}\sqrt{\frac{L}{g_{\mathrm{Mars}}}}}{\frac{1}{2\pi}\sqrt{\frac{L}{g_{\mathrm{Earth}}}}}
\displaystyle= \displaystyle\sqrt{\frac{g_{\mathrm{Earth}}}{g_{\mathrm{Mars}}}}
\displaystyle= \displaystyle\sqrt{\frac{g_{\mathrm{Earth}}}{(0.37)g_{\mathrm{Earth}}}}
\displaystyle T_{\mathrm{Mars}} \displaystyle= \displaystyle\frac{1}{\sqrt{0.37}}T_{\mathrm{Earth}}
\displaystyle= \displaystyle\frac{1}{\sqrt{0.37}}\left(0.80\;\mathrm{s}\right)
\displaystyle= \displaystyle 1.3\;\mathrm{s}

§ Chapter 11, Problem 33

Your grandfather clock’s pendulum has a length of 0.9930\;\mathrm{m}. If the clock loses half a minute per day, how should you adjust the length of the pendulum?

This is a fairly challenging problem. As a starting point, figure out the fractional error.

\displaystyle\frac{30\;\mathrm{s}}{\mathrm{day}}\times\frac{1\;\mathrm{day}}{24\;\mathrm{h}}\times\frac{1\;\mathrm{h}}{3600\;\mathrm{s}} \displaystyle= \displaystyle\frac{1}{2880}

So the clock is off by one part in 2880. We need to shorten the period by this amount such that enough oscillations fit in a 24 hour period and the clock does not lose time.

\displaystyle T_{\mathrm{correct}} \displaystyle= \displaystyle T_{\mathrm{slow}}-\frac{1}{2880}T_{\mathrm{slow}}
\displaystyle= \displaystyle\frac{2879}{2880}T_{\mathrm{slow}}
\displaystyle\frac{T_{\mathrm{correct}}}{T_{\mathrm{slow}}} \displaystyle= \displaystyle\frac{2879}{2880}

Using a ratio

\displaystyle\frac{T_{\mathrm{correct}}}{T_{\mathrm{slow}}} \displaystyle= \displaystyle\frac{\frac{1}{2\pi}\sqrt{\frac{L_{\mathrm{correct}}}{g}}}{\frac{1}{2\pi}\sqrt{\frac{L_{\mathrm{slow}}}{g}}}
\displaystyle\frac{T_{\mathrm{correct}}}{T_{\mathrm{slow}}} \displaystyle= \displaystyle\sqrt{\frac{L_{\mathrm{correct}}}{L_{\mathrm{slow}}}}
\displaystyle L_{\mathrm{correct}} \displaystyle= \displaystyle L_{\mathrm{slow}}\left(\frac{T_{\mathrm{correct}}}{T_{\mathrm{slow}}}\right)^{2}
\displaystyle= \displaystyle 0.9930\;\mathrm{m}\left(\frac{2879}{2880}\right)^{2}
\displaystyle= \displaystyle 0.9923\;\mathrm{m}

Finally

\displaystyle\Delta L \displaystyle= \displaystyle L_{\mathrm{correct}}-L_{\mathrm{slow}}
\displaystyle= \displaystyle 0.9923\;\mathrm{m}-0.9930\;\mathrm{m}
\displaystyle= \displaystyle-0.0007\;\mathrm{m}
\displaystyle= \displaystyle-0.7\;\mathrm{mm}

§ Chapter 11, Problem 36

A fisherman notices that wave crests pass the bow of his anchored boat every 3.0\;\mathrm{s}. He measures the distance between two crests to be 6.5\;\mathrm{m}. How fast are the waves traveling?

\displaystyle v \displaystyle= \displaystyle\lambda f
\displaystyle= \displaystyle(6.5\;\mathrm{m})\frac{1}{3.0\;\mathrm{s}}
\displaystyle= \displaystyle 2.\underline{1}7\frac{\mathrm{m}}{\mathrm{s}}

§ Chapter 11, Problem 37

A sound wave in the air has a frequency of 262\;\mathrm{Hz} and travels with a speed of 343\;\mathrm{m}/\mathrm{s}. How far apart are the wave crests (compressions)?

\displaystyle\lambda \displaystyle= \displaystyle\frac{v}{f}
\displaystyle= \displaystyle\frac{343\frac{\mathrm{m}}{\mathrm{s}}}{262\;\mathrm{Hz}}
\displaystyle= \displaystyle 1.31\;\mathrm{m}

§ Chapter 11, Problem 41

A cord of mass 0.65\;\mathrm{kg} is stretched between two supports 28\;\mathrm{m} apart. If the tension in the cord is 150\;\mathrm{N}, how long will it take a pulse to travel from support to the other?

First we need to calculate the velocity of the waves.

\displaystyle v \displaystyle= \displaystyle\sqrt{\frac{T}{m/L}}
\displaystyle= \displaystyle\sqrt{\frac{150\;\mathrm{N}}{(0.65\;\mathrm{kg})/(28\;\mathrm{m})}}
\displaystyle= \displaystyle 80.\frac{\mathrm{m}}{\mathrm{s}}

We may then calculate the time as

\displaystyle\Delta t \displaystyle= \displaystyle\frac{28\;\mathrm{m}}{80.\frac{\mathrm{m}}{\mathrm{s}}}
\displaystyle= \displaystyle 0.35\;\mathrm{s}

 

Call Me Dr Rob logo