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3 years ago

When the particles of a medium move with simple harmonic motion, this means the wave is necessarily a __________.

Physics

1 answer:

lukranit [14]3 years ago

7 0

Answer:

a. sinusoidal wave

Explanation:

As we know that the medium particle will move in SHM when disturbance travel through the medium

Here the wave propagation through many ways are possible through the medium

When medium particles performs SHM then in that case the wave equation is given as

$y = A sin(\omega t + k x + \phi_0)$

so we have above equation of wave must be sinusoidal equations

so correct answer will be

a. sinusoidal wave

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Read 2 more answers

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top

Anton [14]

Answer:

$y_{max} = 0.829\,m$

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

$v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}$

$v = 2.913\,\frac{m}{s}$

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

$(3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}$

After some algebraic handling, a second-order polynomial is formed:

$12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s$

$1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0$

The roots of the polynomial are, respectively:

$\Delta s_{1} \approx 0.128\,m$

$\Delta s_{2} \approx -0.099\,m$

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

$\Delta s \approx 0.128\,m$

The maximum height that the block reaches after rebound is:

$(3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}$