Pho.rs: Coils of a weighty spring

A1 ^0.40 Count the total number of coils (turns) $N$ of the Slinky spring.

Perform a direct measurement of the number of turns three times. Measurement results: $N_1=N_2=N_3=84$. Average the results and obtain the final answer.

Ответ: $N=84$

A2 ^0.20 Determine the mass $M$ of the entire Slinky spring.

Perform a single direct measurement of the mass on the scale. Obtain the answer.

Ответ: $M=193.3~\text{g}$

A3 ^2.00 Secure the spring in a stand clamp by its top turn. Measure the dependance $L_n$ on $n$ over the widest possible range. Perform at least 20 measurements.

Ответ:

$n$	$l$, mm	$n$	$l$, mm	$n$	$l$, mm	$n$	$l$, mm	$n$	$l$, mm
2	52	20	41	38	29	56	17	74	6
4	54	22	42	40	28	58	16	76	5
6	50	24	40	42	26	60	14	78	3
8	51	26	40	44	26	62	13	80	2
10	49	28	36	46	25	64	13	82	2
12	47	30	35	48	23	66	12
14	46	32	32	50	22	68	10
16	44	34	30	52	21	70	9
18	43	36	30	54	18	72	7

A4 ^0.60 Plot the graph of $L_n$ versus $n$.

Ответ:

A5 ^0.40 Sketch the picture and indicate the external forces acting on this system of bodies.

Ответ: External forces acting on the system of bodies: gravity (directed downwards), and the elastic force acting on the $n$-th turn from the remaining part of the spring (directed downwards).

A6 ^0.40 Using the equilibrium condition, express $l_n$ in terms of $m_0$, $g$, $n$, $N$, and $k$.

The system of bodies consists of $N-(n-1)=N-n+1$ turns. According to the equilibrium equation, $kl_n=(N-n+1)m_0g$, whence $l_n=(N-n+1)m_0g/k$.

Ответ: $l_n=(N-n+1)m_0g/k$

A7 ^0.40 Express $L_n$ in terms of $m_0$, $g$, $n$, $N$, and $k$.

By definition, $L_n=l_n+l_{n+1}=(2N-2n+1)m_0g/k$.

Ответ: $L_n=(2N-2n+1)m_0g/k$

A8 ^0.60 Using the graph plotted in question A4, as well as the measurement results in questions A1 and A2, determine the stiffness coefficient $k$ of a single turn.

The slope of the graph, based on the previous question, is equal to $\alpha=-2m_0g/k$. The value found from the graph is $\alpha=-0.66\text{ mm}$. Thus,
$$k=\frac{-2m_0g}{\alpha}=\frac{-2Mg}{N\alpha}=68.3 \text{ N/m}.$$

Ответ: $k=68.3 \text{ N/m}$

B1 ^2.20 Measure the dependence of the scale readings $m$ on $H$. Perform at least 11 measurements.

Ответ:

$l$, mm	$m$, g	$l$, mm	$m$, g	$l$, mm	$m$, g
298	85.6	214	99.9	114	120.6
291	87.0	202	102.1	100	123.9
283	88.2	191	104.4	90	126.6
275	89.8	180	106.5	77	130.5
263	91.8	171	108.2	66	134.2
250	93.3	161	110.1	56	137.4
241	94.9	148	113.0	44	141.7
233	96.5	137	115.3	34	147.4
224	98.1	126	117.9	20	153.5

B2 ^0.60 Using the result of question A6, express $H$ in terms of $X$, $m_0$, $g$, and $k$. Note that the number of deformed turns is $X$, not $N$.

Use the formula provided in the problem statement:
$$H=\sum_{n=1}^X l_n=\sum_{n=1}^X (X-n+1)\frac{m_0g}{k} = \frac{X+(X-X+1)}{2}\cdot X\cdot\frac{m_0g}{k}.$$

Ответ: $H=\frac{X(X+1)}{2}\cdot\frac{m_0g}{k}$

B3 ^0.40 Sketch the picture and indicate the external forces acting on this system of bodies. What is the elastic force acting on the system?

Ответ: External forces acting on the system of bodies: gravity (directed downwards), and the normal reaction force exerted by the scale (directed upwards). The elastic force from the remaining part of the spring is zero.

B4 ^0.40 Express the scale readings $m$ in terms of $M$, $X$, $m_0$ and $g$.

According to the equilibrium law, $(N-X)m_0g=mg$. Here, the right-hand side represents the normal reaction force exerted by the scale, expressed in terms of its readings. Thus,

Ответ: $m=M-Xm_0$

B5 ^0.40 Using the results of questions B2 and B4, propose a linearization of the dependence $m(H)$.

Based on the answer to question B2,
$$H=\frac{X(X+1)}{2}\cdot\frac{m_0g}{k}\simeq \frac{X^2m_0g}{2k}=\frac{(M-m)^2g}{2m_0k}.$$

Therefore, the required linearization is $H$ versus $(M-m)^2$.

Note. One can also use the linearization of $\sqrt{H}$ versus $m$.

Ответ: $H$ versus $(M-m)^2$, $\sqrt{H}$ versus $m$, or similar.

B6 ^0.60 Plot the linearized graph of the dependence $m$ versus $H$.

Ответ:

B7 ^0.40 Using the plotted graph, determine the stiffness coefficient $k$ of a single turn.

The slope of the linearized graph is $\beta = 28.4 \frac{\text{mm}}{10^3\text{ g}^2}$. On the other hand, based on question B5, $\beta = \frac{1}{2m_0gk}$. Thus,
$$k=\frac{Ng}{2M\beta}=75.0\text{ N/m}.$$

Ответ: $k=75.0\text{ N/m}$

Coils of a weighty spring

Решение