Show work and units Analyzing Newton\'s 2\'lu Law for a mass spring system, we f

Question

Show work and units

Analyzing Newton's 2'lu Law for a mass spring system, we found ax = - k / m x. Comparing this to the x-component of uniform circular motion, we found as a possible solution for the above equation: x = A cos(omega t) vx = - omega A sin(omega t) ax= - omega^2 A cos(omega t) with omega = squareroot k / m. and A the amplitude (maximum displacement from equilibrium). Consider an oscillator with mass 0.365 kg and spring constant 18 Wm. If the amplitude of the oscillation is 4.07 cm, what is the maximum speed?(you might already have solved similar problems employing conservation of energy. Here, I'd like you to find the answer using the equations of motion.) Answer in m/s. Consider an oscillator with frequency 15 Hz. If at t=0 the oscillator is at a maximum displacement from the equilibrium of +5.59 cm, what is the displacement 2.77 seconds later? Answer in cm with the proper sign. Consider an oscillator with mass 0.27 kg and spring constant 18 N/kg. The oscillator is displaced from the equilibrium position by +5.8 cm and released. After what time, is the oscillator for the first time at a displacement of -2.9 cm?

Explanation / Answer

thevelocity will be maximum when sine value is maximum

so, vx = - * A

=> vx= - root( k/m) * A

=> vx= root ( 18/0.365) * ( 0.0407)

= vx= 0.2858 m/s

2) we use x = A cos( * t)

=> x = 5.59 * cos( (2*pi *f * 2.77)

= x= -5.316 cm

3) A= 5.8 cm

x= -2.9 cm

now , x = A cos( * t)

=>-2.9 cm = 5.8 cm * cos( root ( 18/0.27) * t )

=> 2.9/ 5.8 = cos( root ( 18/0.27) * t )

=> arc cos (0.5) = root ( 18/0.27) * t

=> pi/ 3 = 8.165 * t

=> t= 0.128 secs

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.