# Physics 212, Quiz #6b: Jan 5, 1998

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Mozart lifts the lid on his grand piano to tune it. He strikes vertically one particular wire with a mallet, and watches as a wave travels down the wire.

Question 1: Is this a transverse or longitudinal wave?

```  Answer: transverse, since the displacement (vertical) is perpendicular
to the direction of motion (horizontal)
```

The wire is L = 0.909 meters long, and has mass m = 1 gram.

Question 2: In order to yield a wave velocity v = 600 m/s, how tightly must the wire be stretched?

```  Answer: Recall
F
v = sqrt(---)
m/L

solve for F

v^2*m    (600 m/s)^2 * (0.001 kg)
F = ------ = -------------------------
L             0.909 m

= 396 N
```

After the wave moves all the way down the wire, Mozart pulls out his Polaroid camera and takes a snapshot of the wire. Mozart sees that the ends (which are fixed to the piano frame) are at their rest positions, and that there are three other places in the wire, equally spaced from the ends, which are at the rest position, too.

Question 3: What is the wavelength of the wave in the wire? (Hint: draw a picture of Mozart's snapshot)

```  Answer: the wave makes two complete cycles from one fixed end of the
wire to the other.  Therefore, the wavelength is

lambda(wire) = 0.5*(0.909 m) = 0.455 m
```

Mozart recognizes the tone as an A, with a frequency of 1320 Hz. The speed of sound in air is v(air) = 340 m/s.

Question 4: What is the wavelength of the sound wave entering Mozart's ear?

```  Answer: The velocity of a wave is

v = f*lambda

Hence, we can solve for wavelength lambda in air

v     340 m/s
lambda(air) = --- = ---------- = 0.258 m
f     1320 Hz
```

The maximum displacement of the wire from its rest position is 6 mm.

Question 5: Write an equation which describes the displacement "y" of any point in the wire from its rest position, as a function of distance "x" from the left-hand end of the wire.

```  Answer: the general form of the equation for displacement as a function
of position is
2*pi*x                  2*pi*x
y = A * sin(-------)    or  A * cos(------)
lambda                 lambda

In this case, the displacement "y" = 0 at the fixed
end of the wire, where "x" = 0; therefore, we must use
the sin function, because cos(0) = 1.

In this case,

A = 6 mm = 0.006 m         is the amplitude
lambda = 0.455 m                is the wavelength of
the wave in the wire

One could write the above "sin" equation, and then define
the terms "A" and "lambda" (as I have done), or one could
write explicitly
2*pi*x
y = 0.006 m * sin(-------)  = 0.006 m * sin(13.8*x)
0.455 m

where "x" is in meters.
```