INTRODUCTORY PHYSICS  TEST NO. 4  SPRING 08   NAME______________________

     1. A plucked guitar string vibrates in its fundamental mode (that is,
	with one antinode in the middle of the string).  The antinode vibrates
	a total distance of 0.8 cm.  The guitar string is 60 cm long and
	produces a middle C note (512 Hz).  What are the amplitude, wave-
	length, period, and wave speed of the standing wave?

	The total vibrational displacement of the antinode is the peak-to-peak
	amplitude.  The actual amplitude is half the peak-to-peak amplitude
	for a harmonic wave.  Hence

	A = 0.4 cm.

	The node-to-node distance is half a wavelength.  The ends of the
	guitar string are nodes with no node in between.  Hence,

	λ = 2 × 60 cm = 120 cm.

	The period is one over the frequency.  That is,

	T = 1 / f = 1 / 512 Hz = 0.00195 s = 1.95 ms

	The wavespeed is given by the equation

	v = λf = (1.2 m)(512 Hz) = 614 m/s.



     2. If 35 MHz ultrasound is used for a scan of the eye, and the speed of
	the wave is 1150 m/s, how small an object could be detected in the
	eye?

	The smallest object that can be detected will be about the size of a
	wavelength.

	λ = v / f = (1150 m/s) / (35 × 106 Hz) = 3.3 × 10-5 m = 33 μm.


     3. A source of sound radiates equally in all directions with a power of
	126 milliwatts.  (a) What is the intensity in watts per square cen-
	timeter at a distance of 10 m from the source?  (b) How many decibels
	is this above threshold?  (No need to use the logarithm formula unless
	you want to.  Rounding the intensity to the nearest power of 10 and
	then estimating to the nearest 10 dB is fine.)

	The formula for intensity when a wave is radiated equally in all
	directions is

	I = P / 4πr2 = (0.126 W) / 4π(1000 cm)2 = 1.00 × 10-8 W/cm2.

	Threshold is 10-16 W/cm2 and the answer above is
	eight powers of ten more than that.  Each power of ten is 10 dB, so
	the sound is 80 dB above threshold.

	Or, you can use the formula,

	β = 10 log(I / Io) = 10 log (10-8 W/cm2 / 10-16 W/cm2) = 10 log(108) = 80 dB.


     4. A hand-held slide viewer is projecting the image of an illuminated
	slide on a small screen 25 cm on the other side of the lens from the
	slide.  Measurement shows that the slide is 5 cm from the lens.  What
	is the focal length of the lens?  What is its strength in diopters?
	By how much is the slide magnified?

	In the thin-lens formula,

	1/f = 1/O + 1/I,

	O = 5 cm and I = 25 cm.  (I is positive since the fact the image is on
	a screen means it is a real image.)  Put in the numbers and solve for
	f.

	1/f = 1/(5 cm) + 1/(25 cm) = 5/(25 cm) + 1/(25 cm) = 6/(25 cm).

	f = (25 cm)/6 = 4.2 cm.

	S in diopters = 1/f where f is in meters.

	S = 1/(0.042 m) = 24 diopters

	The magnification is given by

	M = - I/O = - (25 cm)/(5 cm) = -5.

	The negative sign says it is an upside down image.  The image is
	therefore five times the size of the object.


     5. Where would you place an object with respect to a lens with a focal
	length of -25 cm in order to get a virtual image 20 cm from the lens?
	(Recall your sign conventions.)

	The negative focal length means you have a diverging lens.  A diverging
	lens can only produce a virtual image, for which the image distance is
	negative by sign convention.  Hence I = -20 cm.  Using the thin lens
	formula,

	1/f = 1/O + 1/I  →  1/(-25 cm) = 1/O + 1/(-20 cm).

	Rearranging,

	1/O = 1/(-25 cm) - 1/(-20 cm) = -4/(100 cm) + 5/(100 cm) = 1/(100 cm).

	Hence O = 100 cm, which is how far from the lens you have to place
	the object.