Question

3. A 250 g mass is attached to a horizontal spring and oscillates with a frequency of 2.1 Hz. At one instant the mass is at -4.3 cm and has a horizontal velocity of 25 cm/s. A. What is the spring constant? B. What is the total energy of the oscillator? C. What is the period of oscillation? D. What is the amplitude? E. What is the maximum speed?

Answer 1

Given that

m = 250 g = 0.250 kg, f = 2.1 Hz, I = -4.3 cm = -0.043 m, v = 25 cm/s = 0.25 m/s

A) Time period

= k = 47²f2m

Spring constant

k = 47?f-m = 47°(2.1) (0.250) = 43.525 43.5 N/m

B)

Total energy of oscillator

E = 5k2? +5 mu?

E = +(43.525)(-0.043)2 + 3(0.250)(0.25)2 = 0.04805 0.048 J

C)

Period of oscillation

可2.1 ll. = 0.4762 <0.48s

D)

Maximum energy of oscillator is
Emar = 24
, this should be equal to total energy of oscillator.
Emar = -k4? = E

2E A = Vă=v 2 * 0.04805 4359 = 0.047 m = 4.7 cm

E)

Maximum speed

V = Aw = A(27f) = 0.047(27 * 2.1) = 0.620 m/s = 62 cm/s