The value of maximum possible amplitude in the case of forced oscillations when driving frequency is close to natural frequency, is

Question

The value of maximum possible amplitude in the case of forced oscillations when driving frequency is close to natural frequency, is (A) (F0/m(ω2-ωd2)) (B) (F0/ωd b) (C) (F0/m ω2) (D) (-F0/m ωd2)

Tardigrade Admin · Accepted Answer

The amplitude of forced oscillation, A=(F0/ m2(ω2-ωd2)2+ωd2 b2 1 / 2) when driving frequency ωd is close to natural frequency ω, so we can take (ωd=ω). Hence, A=(F0/ωd b).

Q. The value of maximum possible amplitude in the case of forced oscillations when driving frequency is close to natural frequency, is

$\frac{F _{0}}{m ( ω ^{2} - ω _{d}^{2} )}$

$\frac{F _{0}}{ω _{d} b}$

$\frac{F _{0}}{m ω ^{2}}$

$\frac{- F _{0}}{m ω _{d}^{2}}$

The amplitude of forced oscillation,
$A = \frac{F _{0}}{{ m ^{2} ( ω ^{2} - ω _{d}^{2} ) ^{2} + ω _{d}^{2} b ^{2} } ^{1/2}}$
when driving frequency $ω_{d}$ is close to natural frequency $ω$ , so we can take $(ω_{d} = ω)$ .
Hence, $A = \frac{F _{0}}{ω _{d} b}$ .

Q. The value of maximum possible amplitude in the case of forced oscillations when driving frequency is close to natural frequency, is

m(ω2−ωd2​)F0​​

ωd​bF0​​

mω2F0​​

mωd2​−F0​​

The amplitude of forced oscillation, A={m2(ω2−ωd2​)2+ωd2​b2}1/2F0​​ when driving frequency ωd​ is close to natural frequency ω, so we can take (ωd​=ω). Hence, A=ωd​bF0​​.

$\frac{F _{0}}{m ( ω ^{2} - ω _{d}^{2} )}$

$\frac{F _{0}}{ω _{d} b}$

$\frac{F _{0}}{m ω ^{2}}$

$\frac{- F _{0}}{m ω _{d}^{2}}$

The amplitude of forced oscillation,
$A = \frac{F _{0}}{{ m ^{2} ( ω ^{2} - ω _{d}^{2} ) ^{2} + ω _{d}^{2} b ^{2} } ^{1/2}}$
when driving frequency $ω_{d}$ is close to natural frequency $ω$ , so we can take $(ω_{d} = ω)$ .
Hence, $A = \frac{F _{0}}{ω _{d} b}$ .