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3 years ago

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A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a

harmonic driving force of frequency 10.5 Hz, resulting in a forced-motion amplitude of 3.00 cm. Determine the maximum value of the driving force..

1 answer:

Nimfa-mama [501]3 years ago

6 0

Answer:

$F_0 = 393 N$

Explanation:

As we know that amplitude of forced oscillation is given as

$A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}$

here we know that natural frequency of the oscillation is given as

$\omega_0 = \sqrt{\frac{k}{m}}$

here mass of the object is given as

$m = \frac{W}{g}$

$\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}$

$\omega_0 = 8.48 rad/s$

angular frequency of applied force is given as

$\omega = 2\pi f$

$\omega = 2\pi(10.5) = 65.97 rad/s$

now we have

$0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}$

$F_0 = 393 N$

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A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

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Explanation:

From the question we are told that

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The initial position of the projectile is s= 1.5m

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Now for the limit

a = 200 m/s

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$\int\limits^{200}_{0} vdv = \int\limits^{3}_{1.5} \frac{c}{s} ds$

$[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.$

$\frac{200^2}{2} = c ln[\frac{3}{1.5} ]$

=> $c = \frac{20000}{0.69315}$

$c = 28853.78 \ m^2 /s^2$

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