91影视

Formula for displacement oscillation is

Equation$1$

$x\left(t\right)=A{e}^{鈭�bt/2m}\cos 鈦�(蝇t+蠒)$

Damped oscillation has a equation of

Equation$2$

$蝇=\sqrt{\frac{g}{L}鈭�\frac{b^2}{4m^2}}$

Dampled oscillation is varied in time

Equation $3$

$E\left(t\right)=E_0{e}^{鈭�bt/m}$

Terms of frequency in a wave

Equation 4

$蝇=2蟺f$

Formula for gravitation potential energy

Equation $5$

$U_g=mgy$

Equation ($2$) yields the frequency content of the decelerated oscillator:

$蝇=\sqrt{\frac{g}{L}鈭�\frac{b^2}{4m^2}}$

Fill up the blanks with numerical values from the given data:

$蝇=\sqrt{\frac{9.80\mathrm{m}/{\mathrm{s}}^2}{0.45\mathrm{m}}鈭�\frac{(0.010\mathrm{kg}/\mathrm{m}{)}^2}{4(0.350\mathrm{kg}{)}^2}}$

$=4.666644801{\mathrm{s}}^{鈭�1}$

Equation ($4$) yields the phase of the damped oscillator:

$T=\frac{2蟺}{蝇}$

$=\frac{2蟺}{4.666644801{\mathrm{s}}^{鈭�1}}$

$=1.346s$

The oscillator obtained the required number of oscillations in $25s$:

$n=\frac{t}{T}$

$=\frac{25\mathrm{s}}{1.346\mathrm{s}}$

$=18.6$oscillations

Allow the string's gravitational potential energy to be 0 at its lowest point. Equation ($5$) shows that the string possesses a gravitational potential energy at first:

$E_0=U_g=mgh$

The height is calculated using $胃$, the angle, and $L$, the string length. As a result, $h-L(1-\cos 胃)$

$E_0=mgL(1鈭�\cos 鈦�胃)$

Equation ($3$) expresses the string's energy as a function of time:

$E\left(t\right)=E_0{e}^{鈭�bt/m}$

The heat absorbed by that of the string after $25$is equal to the string's original energy minus the stage's energy at time $25$.$t=25s$So

$E_{\mathrm{lost}}=E_0鈭�E\left(t\right)$

$\begin{array}{r}=E_0鈭�E_0{e}^{鈭�bt/m}\\ =E_0\left(1鈭�e^{鈭�bt/m}\right)\\ \end{array}$

$=mgL(1鈭�\cos 鈦�胃)\left(1鈭�e^{鈭�bt/m}\right)$

Numericals values are substituted

role="math" localid="1649966946821" $$\begin{gathered} E_{\text{lost}}=\left(0.350\mathrm{kg}\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(0.45\mathrm{m}\right)\left(1鈭�\cos 鈦�{4.5}^{鈭�}\right), \\ \left(1鈭�\exp 鈭�\frac{(0.010\mathrm{kg}/\mathrm{s})\left(25\mathrm{s}\right)}{0.350\mathrm{kg}}\right) \end{gathered}$$

$=2.4脳{10}^{鈭�3}\mathrm{J}$