A 2.5 kg mass is at rest hanging vertically attached to aspring that has a sprin

Question

A 2.5 kg mass is at rest hanging vertically attached to aspring that has a spring constant, k = 3300 N/m. This mass isstruck by a hammer that transfers 5.0 J of kinetic energy to themass and causes it to oscillate up and down as a one dimensionalharmonic oscillator. A.What is the amplitude of the simple harmonic motion? B. What is the frequency of the simple harmonice motion? C. What is the speed of the mass when x is 0.03 m below theequilibrium point? D. Write an equation the describes y as a function of t. A 2.5 kg mass is at rest hanging vertically attached to aspring that has a spring constant, k = 3300 N/m. This mass isstruck by a hammer that transfers 5.0 J of kinetic energy to themass and causes it to oscillate up and down as a one dimensionalharmonic oscillator. A.What is the amplitude of the simple harmonic motion? B. What is the frequency of the simple harmonice motion? C. What is the speed of the mass when x is 0.03 m below theequilibrium point? D. Write an equation the describes y as a function of t. A.What is the amplitude of the simple harmonic motion? B. What is the frequency of the simple harmonice motion? C. What is the speed of the mass when x is 0.03 m below theequilibrium point? D. Write an equation the describes y as a function of t.

Explanation / Answer

   Given         mass  m   = 2.5 kg spring constant k = 3300 N /m       Kinetic energy  E = 5 J A)     Amplitude :             we know that                     E = (1/2)k A2                         5J = 0.5 x 3300 N /M x A2                          A = 5 / ( 0.5 x 3300 )                              = 0.055 mm B) Frequency    :                 f   =   ( 1 / 2 ) k / m                     = ( 1 / 2 x 3.14 ) ( 3300 / 2.5                      = 5.78 Hz C )   Energy   E = mgh + (1/2) m v 2                  5 = (2.5 x 9.8 x 0.03 )+ (0.5 x 2.5x v2                     v   = 1.8747 m/s D)    y = A sin ( t )             = 0.055 sin (2f t )             = 0.055 sin (36.2984 t )                          A = 5 / ( 0.5 x 3300 )                              = 0.055 mm B) Frequency    :                 f   =   ( 1 / 2 ) k / m                     = ( 1 / 2 x 3.14 ) ( 3300 / 2.5                      = 5.78 Hz C )   Energy   E = mgh + (1/2) m v 2                  5 = (2.5 x 9.8 x 0.03 )+ (0.5 x 2.5x v2                     v   = 1.8747 m/s D)    y = A sin ( t )             = 0.055 sin (2f t )             = 0.055 sin (36.2984 t )

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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