Which wave has the smallest amplitude?

Physics

1 answer:

nydimaria [60]3 years ago

8 0

Answer:

C. C

Explanation:

A wave can be defined as a disturbance in a medium that progressively transports energy from a source location to another location without the transportation of matter.

In Science, there are two (2) types of wave and these include;

I. Electromagnetic waves: it doesn't require a medium for its propagation and as such can travel through an empty space or vacuum. An example of an electromagnetic wave is light.

II. Mechanical waves: it requires a medium for its propagation and as such can't travel through an empty space or vacuum. An example of a mechanical wave is sound.

A crest can be defined as the highest (vertically) point on a waveform.

On a related note, a trough is the lowest (vertically) on a waveform.

An amplitude can be defined as a waveform that's measured from the center line (its origin or equilibrium position) to the bottom of a trough or top of a crest. Thus, the vertical axis (y-axis) is the amplitude of a waveform i.e it's measured vertically.

In this scenario, waveform C which is represented by a blue curvy line has the smallest amplitude in comparison with the other waveforms because it has the minimum height when measured from the origin.

In contrast, waveform A represented by a purple line has the highest amplitude because it has the maximum height when measured from the origin.

Mathematically, the amplitude of a wave is given by the formula;

x = Asin(ωt + ϕ)

Where;

x is displacement of the wave measured in meters.
A is the amplitude.
ω is the angular frequency measured in rad/s.
t is the time period measured in seconds.
ϕ is the phase angle.

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You have a pendulum clock made from a uniform rod of mass M and length L pivoting around one end of the rod. Its frequency is 1

drek231 [11]

The new oscillation frequency of the pendulum clock is 1.14 rad/s.

The given parameters;

Mass of the pendulum, = M
Length of the pendulum, = L
Initial angular speed, $\omega _i$ = 1 rad/s

The moment of inertia of the rod about the end is given as;

$I_i = \frac{1}{3} ML^2$

The moment of inertia of the rod between the middle and the end is calculated as;

$I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} = \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}$

Apply the principle of conservation of angular momentum as shown below;

$I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f$

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

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3 years ago

A 250 g object is hung onto a spring. It stretches 18 cm. Find the spring's spring constant

Juliette [100K]

<h2>Answer: 13.61 N/m</h2>

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force $F$ applied to it, as long as the spring is not permanently deformed: