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INPhO 2022, Теория — Задача 1

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Indian Olympiad Qualifier in Physics (IOQP) 2021-2022 — Part II: Indian National Physics Olympiad (INPhO) · 13 март 2022 г. · 12 т.

Указания

This booklet consists of 4 pages and total of 5 questions. Write roll number at the top wherever asked.
Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.
Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the Detailed Answer Sheets in the Answer Booklet. Simple short answers and plots may be directly entered in the Summary Answer Sheet. Marks may be deducted for absence of detailed work in questions involving longer calculations.
Strike out any rough work that you do not want to be considered for evaluation. You may also use the space on the Question Paper for rough work – this will NOT be evaluated.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Please submit the Answer Booklet at the end of the examination. You may retain the Question Paper.

Условие

A block of mass $m = 0.1\,\mathrm{kg}$ is attached to a spring (one end fixed to the wall) with spring constant $k = 50\,\mathrm{N\,m^{-1}}$ . The block slides on a rough horizontal table along the $x$ -axis. Assume that both the coefficients of kinetic ( $\mu_k$ ) and static friction ( $\mu_s$ ) are same and constant ( $\mu_k = \mu_s = \mu = 0.25$ ). The block is initially displaced to $x_0 = 0.1\,\mathrm{m}$ from the unstretched position (normal length of the spring, $x = 0$ ) of the spring and released from rest as shown below. Neglect any air resistance. Take the acceleration $g$ due to gravity to be $10\,\mathrm{m/s^2}$ .

A block attached to a horizontal spring whose left end is fixed to a wall; a vertical dashed line marks the unstretched position, and an arrow from it to the block is labelled x0.

(a) How many times ( $n$ ) will the block cross the unstretched position before coming to rest permanently? [3 т.]

(b) Determine the total distance $D$ covered by the block before coming to rest. [1 т.]

(c) Let us divide one complete oscillation of the block, starting from a fully stretched condition of the spring, into four distinct sections, requiring the following times in order:

(i) $t_1$ : time taken for the block to move from fully stretched to the unstretched position,

(ii) $t_2$ : time taken for the block to move from the unstretched position to fully compressed position,

(iii) $t_3$ : time taken for the block to move from fully compressed to the unstretched position,

(iv) $t_4$ : time taken for the block to move from the unstretched position to fully stretched position.

Let the distance covered during the above intervals be $d_1$ , $d_2$ , $d_3$ , and $d_4$ , respectively.

Also, let $T_1$ and $T_2$ be the time taken to complete the first and the second oscillations, respectively, starting from the initial displacement, $x_0$ .

Compare the above times and distances by inserting an appropriate sign (from among $<$ , $>$ , or $=$ only) between the given quantities in each of the boxes below. Note that you will be penalised for 0.5 marks for giving each incorrect answer in this part. You need not to justify your answer.

$t_1 \qquad t_2$	$t_2 \qquad t_3$	$t_1 \qquad t_3$
$d_1 \qquad d_2$	$d_2 \qquad d_4$	$T_1 \qquad T_2$

[6 т.]

(d) Qualitatively plot the displacement $x$ from the unstretched position vs the time $t$ . [2 т.]

Отговори

Решение

(a)

Four sketches of the spring-block system against a dashed vertical line marking the unstretched position: the initial stretch x0, then the block on the compressed side at x1, stretched again at x2, and compressed at x3, each distance shown by an arrow.

m\ddot{x} = -kx \pm \mu m g \qquad (1.1)

where the $\pm$ sign is used so that the friction is opposite to the direction of the velocity of the block. Lets define the ratio of the frictional force to the maximum restoring force to be $\alpha$ i.e. $\alpha = \mu m g/k x_0$ . During the first half cycle of the motion, the loss of potential energy is equal to the work done against friction

\frac{k x_0^2}{2} - \frac{k x_1^2}{2} = \mu m g (x_0 - x_1) \qquad (1.2)

where $x_1$ is the displacement after one half cycle. This gives

x_1 = -x_0 + 2\alpha x_0 \qquad (1.3)

x_2 = -x_1 - 2\alpha x_0 \qquad (1.4)

Thus

x_j = -x_{j-1} - (-1)^j 2\alpha x_0 \qquad (1.5)

x_j = (-1)^j (1 - 2\alpha j) x_0 \qquad (1.6)

The block will come to rest permanently at $x_n$ when

|x_n| \le \alpha x_0 < |x_{n-1}| \qquad (1.7)

\frac{1-\alpha}{2\alpha} \le n < \frac{1+\alpha}{2\alpha} \qquad (1.8)

For the given values of $\mu, m, k$ and $x_0$ , $\alpha = 0.05$ . Thus $9.5 \le n < 10.5$ . The block will cross the unstretched position 9 times and then comes to the rest at the unstretched position.

(b) Total distance

D = x_0 + \sum_{j=1}^{n-1} x_j + x_n \qquad (1.9)

= x_0 + \sum_{j=1}^{n-1} (-1)^j (1 - 2\alpha j) x_0 + 0.1 \qquad (1.10)

= 2n(1 - \alpha n) x_0 + 0.1 = 1.00\,\mathrm{m} \qquad (1.11)

(c)

$t_1 > t_2$	$t_2 < t_3$	$t_1 < t_3$
$d_1 > d_2$	$d_2 > d_4$	$T_1 = T_2$

(d) Equations of motion are

m\ddot{x} = -kx + \mu m g \qquad (\dot{x} < 0) \qquad (1.12)

m\ddot{x} = -kx - \mu m g \qquad (\dot{x} > 0) \qquad (1.13)

The general solution to the above equation is simply the solution for the SHO and an additive constant.

x = A\cos\omega t + C \qquad (\dot{x} < 0) \qquad (1.14)

x = A\cos\omega t - C \qquad (\dot{x} > 0) \qquad (1.15)

and velocity and the acceleration are

\dot{x} = -A\omega\sin\omega t \qquad (1.16)

\ddot{x} = A\omega^2\cos\omega t \qquad (1.17)

Using the above equations in Eqs. (1.12) and (1.13)

x(t) = A\cos\omega t + \alpha x_0 \qquad (\dot{x} < 0) \qquad (1.18)

x(t) = A\cos\omega t - \alpha x_0 \qquad (\dot{x} > 0) \qquad (1.19)

For the given value of $\mu$ , $\alpha = 0.05$ . Every half swing exhibits simple harmonic motion. Turning points are regularly spaced in time at every $\pi/\omega$ intervals. The initial position at each turning point (or the end position of the previous half turn) determines the amplitude and phase for the following half turn. During the time interval between the turning point $t_j$ and $t_{j+1}$

x_j(t) = A_j\cos\omega t - \alpha x_0 (-1)^j \qquad (1.20)

Note, for the compression of the spring, $\dot{x} < 0$ and $j = 1, 3, 5....$ Comparing Eqs. (1.20) and (1.6)

(-1)^j (1 - 2\alpha j) x_0 = A_j (-1)^j - \alpha x_0 (-1)^j \qquad (1.21)

A_j = x_0 (1 - \alpha(2j-1)) \qquad (1.22)

To summarize the results, the motion of each half swing is essentially the motion of a simple harmonic oscillator centered about either $\alpha x_0$ or $-\alpha x_0$ . At each turning point ( $t = j\pi/\omega$ ), the amplitude of the oscillator decreases by $2\alpha x_0$ . Before $j = n$ , given by the Eq. (1.8), the mass moves according to

x_j(t) = x_0 \left[\{1 - \alpha(2j-1)\}\cos\omega t - (-1)^j \alpha\right] \qquad (1.23)

The plot of $x$ vs $t$ for the given value of $\alpha$ is shown below. The displacement lies within a pair of straight lines (shown by the dotted lines) with slopes $\pm 2\alpha x_0/(\pi/\omega)$ .

Graph of x (m) from −0.1 to 0.1 against t (s) from 0 to 1.4: a damped oscillation whose peaks lie on two dashed straight lines converging to zero near t = 1.4 s.

Оригинал в Архива: IOQP2022-PartII-Questions-en.pdf · официални решения: IOQP2022-PartII-Solutions-20220503.pdf