IPhO 1997, Експеримент — Задача 1. Characterization of the bimorph

1.1 · Voltage circuit

1 point for any diagram which allows applying the full range of voltages.

1.2 · Geometry

$D$ is the distance from the mirror to the spot on the screen. $y$ is the distance the spot moves when voltage is applied, $x$ is the shortest distance from the mirror to the screen.

1 point for any well-labelled arrangement allowing the experiment to be performed with reasonable sensitivity.

1.3 · Displacement

$l$ is the length of the free part of the bimorph. See Fig. 2 for $x$, $y$ and $D$.

When the bimorph bends into an arc shape, the mirror at the end is rotated by an angle [symbol missing]/2. This results in a change of direction of the reflected beam by an angle [symbol missing]. The spot formed by the reflected beam on the screen moves by a distance $y$ (see Fig. 2). If the angle between the reflected beam and the screen is [symbol missing] (less than 90 degrees to increase accuracy), the small angle [symbol missing] by which the laser beam is deflected due to the movement of the bimorph is given in the source as:

The angle of rotation of the mirror [symbol missing]/2 is related to the deflection $d$ of the free end of the bimorph by $d=l\sin\text{[symbol missing]}/4$ (see Fig. 3). For small [symbol missing] we have

2 points (also for any other correct formula relating the displacement $d$ to the measured quantities). 1 point for the formula with 1 error, 1/2 for the formula with two errors, 0 for the formula with more errors. Total 2 points.

1.4 · Displacement versus voltage

2 points for a correct graph.

1.5 · Dissipated energy

The force moving the bimorph is proportional to the applied voltage $V$, the distance is proportional to the displacement $d$, so the area under the plot of $d$ versus $V$ is proportional to the energy. Therefore the area inside the hysteresis curve is proportional to the energy dissipated in the bimorph.

1 point for identifying the area inside the hysteresis curve as proportional to the energy dissipated in the bimorph.

1.6 · Hysteresis-loop area

Graph 2 shows the data from Graph 1 with the hysteresis curve outlined. Area must be expressed in the units of the x axis times units of the y axis. The area estimated from Graph 2 is printed as $(350\pm160)\,\mathrm{Vm}$; the source does not render the prefix symbol between V and m.

If the quantity is correctly identified and the value consistent to within 50% with the student’s data, 1 point will be given. No points will be allocated for the value if the units are not specified. 1/2 point for a reasonable error estimate. Total 1.5 points.

2.1 · Voltage exponent

If hysteresis is ignored, the data from Graph 1 can be fitted with a straight line, indicating that the displacement is proportional to voltage. See Graph 3.

1 point for showing that $m=1$.

2.2 · Length exponent

Graph 4 shows the dependence of the displacement of the end of the bimorph on the length of the bimorph for the same voltage change, 72 V.

One can conclude from this graph that the displacement is not proportional to the length. One can then plot $\log d$ versus $\log l$ (see Graph 5) and calculate the slope, which will give the value of $n$. From Graph 5 we obtain $n=1.89\pm0.13$. Notice that $l$ should be measured from the centre of the mirror to the edge of the holder.

1 point for demonstrating by correct measurements and graph and/or calculations that $n$ is about 2 (between 1.7 and 2.3). 1/2 point for values between 1.5 and 1.7 or 2.3 and 2.5. 1/2 point for a reasonable error estimate, for example from the range of possible slopes on Graph 5. Total 1.5 points.

Source discrepancy: the printed Graph 5 labels its horizontal axis log(d) and vertical axis log(l), although the accompanying text specifies log(d) versus log(l). The original figure and text are preserved.

2.3 · Constant A

Knowing $l$, $m$ and $n$, one can calculate $A$ from the slope of the straight line on Graph 2.

1 point for a reasonable error estimate, for example from the range of possible slopes on Graph 3. Total 3 points.

Source note: the reference to “Graph 2” above is printed in the original solution.

3.1 · Capacitance circuit

Missing source image: the original screenshot displays a broken image in place of this circuit diagram. No replacement circuit is supplied here.

1 point for this or equivalent schematics utilizing the resistor printed as 1 G [unit symbol missing] in such a way that the time constant is of the order of seconds. 1/2 point for other arrangements that could measure the capacitance. Total 1 point.

3.2 · RC decay

The resistor printed as 1 G [unit symbol missing] and the bimorph form an RC circuit. If we charge the capacitance of the bimorph to maximum voltage ($V_0=36\,\mathrm V$) and disconnect the battery, the voltage will decay according to the formula

where $t$ is elapsed time. If the voltmeter with the input resistance printed as 1 M [unit symbol missing] is connected to the bimorph to measure the voltage directly, the time constant is too short to be measured. The deflection of the bimorph is proportional to the voltage, so one can measure the decay time indirectly by observing the decay of the deflection after the bimorph was charged. To do so efficiently, one can mark several points on the screen between the deflections corresponding to the maximum voltage and 0 voltage and measure the time the deflection takes to decay to these points. The slope of the plot of logarithm of deflection versus time (Graph 6), or the equivalent calculations, will give the time constant; knowing $R$, one can obtain $C$.

1 point for a correct formula for $C$. 1 point for using the time-dependent displacement $d$ as a measure of the time constant. 1 point for a correct data table and/or graph. Total 3 points.

3.3 · Capacitance

No points will be allocated for the value if the units are not specified. 1/2 point for a reasonable error estimate. Total 2 points.