Mechanical Oscillations and Waves

$x = 4 \sin 200\pi t$.

$T/4$; $T/12$; $T/6$.

1st method. The following figure shows the graph of the displacement of the oscillating body versus time. At the initial moment, the displacement is zero. The phases in radians are marked on the abscissa axis. From the graph, it is clear that the entire path from the mean position to the extreme maximum displacement takes $T/4$ for the oscillating body. The oscillating body covers the first half of the path in $T/12$, which corresponds to the phase $\pi/6$ ($\sin(\pi/6) = 0.5$). The second half of the path is covered in time $T/4 - T/12 = T/6$.

2nd method. The displacement $x$ of the body is determined by the formula $x = a \sin(2\pi t / T)$, where $a$ is the amplitude, equal to the displacement to the extreme position; $t$ is the time counted from the moment of passing the equilibrium position. The time $t_1$ required for the body to travel from the mean position to the extreme position is determined from the condition $x = a$: $a = a \sin(2\pi t_1 / T)$, whence $2\pi t_1 / T = \pi/2$ and $t_1 = T/4$. The time $t_2$ required to travel the first half of this path is determined from the condition $x = a/2$, i.e., $a/2 = a \sin(2\pi t_2 / T)$, whence $2\pi t_2 / T = \pi/6$ and $t_2 = T/12$. The time $t_3$ to travel the second half of the path is $t_3 = t_1 - t_2 = T/6$.

$T/3$.

See the solution to the previous problem.

Circular motions of a 'conical' pendulum can be obtained if the deflected pendulum is given a push (velocity) in a direction perpendicular to the plane of possible oscillations after the initial deflection. Thus, 'conical' motions can be considered as the result of the superposition of two independent mutually perpendicular oscillations. The periods of these oscillations are the same (any plane of swing is no different from any other). Consequently, the period of the complex motion - the revolution of the pendulum along the cone - will be the same as the period of oscillation of a simple pendulum. The period of revolution of a 'conical' pendulum performing circular motion at small angles at the apex of the cone is $T = 2\pi \sqrt{l/g}$ (see problem 30 in Chapter: Circular Motion).

Approximately 0.3% longer.

The condition for the equality of the periods of oscillation of pendulums of different lengths at different heights is expressed as follows: $l_0/g_0 = l/g_h$. Forming the derived proportion, we get $\frac{l_0 - l}{l_0} =\frac{g_0 - g_h}{g_0}.$ Considering that $g_h = \frac{\gamma M}{(R+h)^2} \approx g_0 (\frac{R}{R+2h})$ (see problem 5 in Chapter: Conservation of Energy), it is easy to find $\frac{l_0 - l}{l_0} \approx 2h/(R+2h) \approx 2h/R \approx 3 \times 10^{-3}.$

$\Delta t \approx 67.5$ s.

The period of oscillation of a pendulum at height $h$ is $T = 2\pi \sqrt{\frac{l}{g_h}}.$ The acceleration of gravity at this height is $g_h = g_0 (\frac{R}{R+h})^2,$ where $g_0$ is the acceleration at the Earth's surface and $R$ is its radius (see problem 5 in Chapter: Conservation of Energy). Then $T = 2\pi \sqrt{\frac{l}{g_h}} = 2\pi \sqrt{\frac{l}{g_0}} \left(\frac{R+h}{R}\right) = T_0 \frac{R+h}{R} = T_0 \left(1 + \frac{h}{R}\right),$ where $T_0$ is the period of oscillation at the Earth's surface. Thus, over time $T_0$, the clock lags by $\Delta T = T - T_0 = h\frac{T_0}{R}.$ Over time $t = 24$ h, the clock will lag by some time $\Delta t$, which relates to $t$ as $\Delta T$ relates to $T_0$, and obviously, $\Delta t / t = \Delta T / T_0$, whence $\Delta t = t \frac{\Delta T}{T_0} = t\frac{h}{R}.$

For estimation, assume the deposit has the shape of a sphere of radius $r$. Then the force of gravity acting on a body of mass $m$ near the deposit should be less than far from it (on the Earth's surface) by the amount $\frac{\gamma m (\rho_0 - \rho) \cdot 4/3 \pi r^3}{r^2} = \frac{4}{3} \pi \gamma m (\rho_0 - \rho) r$. The acceleration $g'$ of free fall near the deposit will therefore be equal to $g - \frac{4}{3} \pi \gamma (\rho_0 - \rho) r$, and the period of oscillation of the pendulum $T' = 2\pi \sqrt{l/g'} = 2\pi \sqrt{l/g} \sqrt{g/g'} = T \sqrt{g/g'}$, where $T$ is the period of oscillation of the pendulum far from the deposit. Knowing $g'$, it is easy to find $r$.

The period of oscillation of the pendulum will increase.

The action of the buoyant force on the pendulum (Archimedes' force) reduces the tension force of the string, which is equivalent to a decrease in $g$.

$a = g \left[1 - \left(\dfrac{T}{T_1}\right)^2\right] \approx 0.17g$. The acceleration is directed downwards. The direction of motion of the lift does not matter.

When the lift moves with constant acceleration $a$, the tension force of the pendulum string $F_N$ in its equilibrium position relative to the cabin is determined from Newton's second law: $ma = mg - F_N$, whence $F_N = m(g - a)$. In the last formula, $a$ is an algebraic quantity: positive when the lift's acceleration is directed downwards, and negative when the acceleration is directed upwards. It follows that when the pendulum is deflected, the force restoring it to the equilibrium position will be proportional not to $g$, but to $(g - a)$. This means that in the lift moving with acceleration $a$, a pendulum of length $l$ has a period $T_1 = 2\pi \sqrt{l/(g-a)}$. Taking the ratio of the periods of oscillation of the pendulum in the lift moving with acceleration and in a stationary lift and squaring it, we get $\left(\frac{T_1}{T}\right)^2 = \frac{g}{g-a},$ whence we find the answer: $a = g \times \left[1 - \left(\frac{T}{T_1}\right)^2\right] \approx 0.17g;$ the answer is positive, meaning the lift moves with acceleration directed downwards; the direction of velocity does not matter.

$T = 2\pi \sqrt{l/\sqrt{g^2 + a^2}}$.

In the equilibrium position relative to the wagon, the tension force of the pendulum string $F_N$ will be equal to: $F_N = m\sqrt{g^2 + a^2}$.

$\lambda = 7.25$ m.

The tuning fork clamped in the vise sounds longer. The radiation of sound wave energy per unit time from the tuning fork standing on the resonator box is greater.

1 m.

$v \approx 67.5$ km/h.

When the wagon wheels hit the rail joints, the wagon receives an impulse that has both vertical and horizontal components. If the period between impacts is equal to the period of oscillation of the pendulum, the latter will swing particularly strongly.

$\pi/2$.

If two points are separated by a distance equal to $\lambda$, the phase difference is $2\pi$, and if by $l$, then $2\pi l / \lambda$.

When moving along the surface of the sphere.

When moving along the surface of the sphere $t_1 = T/4 = \frac{1}{4} \cdot 2\pi \sqrt{\frac{R}{g}}$ ($T$ is the period of oscillation of a pendulum of length $R$. Explain!). When moving along the inclined plane $t_2 = \sqrt{\frac{2s}{a}} = \sqrt{2\frac{2R \sin \alpha} {g \sin \alpha}} = 2 \sqrt{\frac{R}{g}} > t_1$ ($\alpha$ is the angle of inclination of the plane).

$T = 2\pi \sqrt{\Delta l / g} \approx 0.4$ s.

For series connection of springs $T = 2\pi \sqrt{m\dfrac{k_1+k_2}{k_1 k_2}$. For parallel connection $T = 2\pi \times \sqrt{\dfrac{m}{k_1+k_2}}$.

See problem 14 in Chapter: Statics.