Question

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
m = 250 g = 0.25 kg
f = 2.1 hz
at x = -4.3 cm, v = 25 cm/s

A) we knwo, w = sqrt(k/m)

w^2 = k/m

==> k = m*w^2

= m*(2*pi*f)^2

= 0.25*(2*pi*2.1)^2

= 43.5 N/m

B) Total energy of the oscillator, TE = (1/2)*k*x^2 + (1/2)*m*v^2

= (1/2)*43.5*0.043^2 + (1/2)*0.25*0.25^2

= 0.0480 J

C) T = 1/f

= 1/2.1

= 0.476 s

D) TE = (1/2)*k*A^2

==> A = sqrt(2*TE/k)

= sqrt(2*0.048/43.5)

= 0.0470 m (or) 4.70 cm

E) V_max = A*w

= A*sqrt(k/m)

= 0.047*sqrt(43.5/0.25)

= 0.620 m/s (or) 62.0 cm/s