A steady force of 120 N is required to push | vedclass

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(A) $1$. Calculate the damping constant $b$: Since the force $F = bv$,we have $b = F/v = 120/1 = 120 \ kg/s$.$2$. Calculate the damping parameter $r$:

Vedclass · Accepted Answer

$0.56$

Vedclass · Answer

(A) $1$. Calculate the damping constant $b$: Since the force $F = bv$,we have $b = F/v = 120/1 = 120 \ kg/s$.$2$. Calculate the damping parameter $r$: $r = b / (2m) = 120 / (2 	imes 700) = 120 / 1400 = 6/70 \approx 0.0857 \ s^{-1}$.$3$. Calculate the spring constant $k$: From $F = kx$,we have $k = 450 / 2 = 225 \ N/m$.$4$. Calculate the natural angular frequency $\omega_0$: $\omega_0 = \sqrt{k/m} = \sqrt{225/700} = \sqrt{9/28} \approx 0.5669 \ rad/s$.$5$. Calculate the angular frequency of damped $SHM$ $\omega'$: $\omega' = \sqrt{\omega_0^2 - r^2} = \sqrt{225/700 - (6/70)^2} = \sqrt{22500/70000 - 36/4900} = \sqrt{225/700 - 36/4900} = \sqrt{(1575 - 36)/4900} = \sqrt{1539}/70 \approx 39.23 / 70 \approx 0.56 \ rad/s$.

$A$ steady force of $120 \ N$ is required to push a boat of mass $700 \ kg$ through water at a constant speed of $1 \ m/s$. If the boat is fastened by a spring and held at $2 \ m$ from the equilibrium position by a force of $450 \ N$,find the angular frequency of damped $SHM$ in $rad/s$.

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