Question

A mass of 2 kg is suspended from a vertical spring of stiffness 15 kN/m and subject to viscous damping of 5 Ns/m. What is the amplitude of the forced oscillations produced when a periodic force of amplitude 25 N and angular frequency 100 rad/s acts on the mass? What is the maximum force transmitted to the support of the spring?

Answer 1

Amplitude of A is 4.975 mm and total force is 94.3 N

Step-by-step explanation:

given data in question

mass (m) = 2 kg

stiffness (k) = 15 kN/m

viscous damping (c) = 5Ns/m

amplitude (F) = 25 N

angular frequency (ω) = 100 rad/s

to find out

amplitude of the forced and maximum force transmitted

Solution

static force for transmitted is mg i.e 2 × 9.81 = 19.6 N .............. 1

we know the amplitude formula i.e.

Amplitude of A = amplitude /
`\sqrt{c^(2)\omega^(2) + (k - m \omega^(2))^(2)`

now put the value c k m and ω and we find amplitude

Amplitude of A = 25 /
`\sqrt{5^(2) * 100^(2) + (15000 - 2 * 100^(2))^(2)`

Amplitude of A = 4.975 mm

now in next part we know the maximum force value when amplitude is equal displacement i.e.

maximum force = amplitude of A
`\sqrt{k^(2)+c^(2)\omega^(2)}`

now put all these value c , ω k and amplitude and we get

maximum force = 4.975
`\sqrt{15000^(2)+5^(2) * 100^(2)}`

maximum force = 74.7 N .......................2

total force is combine equation 1 and 2 we get

total force 19.6 + 74.7 = 94.3 N