91影视

Frequency in SHM is

$\mathbf{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{蟿蟿}}\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}$

Period of Time in SHM is

$\mathbf{T}\mathbf{=}\mathbf{2}\mathbf{蟿蟿}\sqrt{\frac{\mathbf{m}}{\mathbf{k}}}$

Angular Frequency in SHM is

$\mathbf{蝇}\mathbf{=}\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}$

(a) When the pebble is suddenly lifted out vertically, this means the mass is decreased. The frequency is related to the mass by

$\mathrm{f}=\frac{1}{2\mathrm{蟿蟿}}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$

As shown by above equation the frequency is inversely proportional to the root square of the mass

$\mathrm{f}鈭�\frac{1}{\sqrt{\mathrm{m}}}$

So, when the mass decreases the frequency increases.

(b) The time period is related to the mass and the force constant by in the form

$\mathrm{T}=2\mathrm{蟿蟿}\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

As shown, the period is proportional to the root square of the mass

$T鈭�\sqrt{m}$

So, when the mass decreases the period decreases.

(c) The amplitude depends on the frequency and the period of the spring inly, not the mass that attached to the spring. So, the amplitude after reducing the mass will be the same

(d) All potential energy converts to kinetic energy. Hence, the kinetic energy is equal to the total mechanical energy in the spring

$\mathrm{K}=\mathrm{E}=\frac{1}{2}{\mathrm{kA}}^2$

As the amplitude A is the same, so the kinetic energy is the same

(e) The angular velocity$蝇$is related to the mass and the force constant by

$\mathrm{蝇}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$

As shown, the angular frequency is inversely proportional to the root square of the mass

$蝇鈭�\frac{1}{\sqrt{\mathrm{m}}}$

So, when the mass decreases the velocity increases.