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

Mass of the object = m = 250 g = 0.25 kg

Spring constant = k

Frequency of oscillation = f = 2.1 Hz

$f = \frac{1}{2\pi }\sqrt{\frac{k}{m}}$

k = 4$\pi$²f²m

k = 4$\pi$²(2.1)²(0.25)

k = 43.525 N/m

Position of the mass = X₁ = -4.3 cm = -0.043 m

Horizontal velocity of the mass = V₁ = 25 cm/s = 0.25 m/s

Total energy of the oscillator = E

E = kX₁²/2 + mV₁²/2

E = (43.525)(-0.043)²/2 + (0.25)(0.25)²/2

E = 0.048 J

Period of oscillation = T

$T = \frac{1}{f}$

$T = \frac{1}{2.1}$

T = 0.476 sec

Amplitude of the motion = A

E = kA²/2

0.048 = (43.525)A²/2

A = 4.7 x 10^-2 m

A = 4.7 cm

Maximum speed of the object = V

E = mV²/2

0.048 = (0.25)V²/2

V = 0.62 m/s

A) Spring constant = 43.525 N/m

B) Total energy of the oscillator = 0.048 J

C) Period of oscillation = 0.476 sec

D) Amplitude of the motion = 4.7 cm

E) Maximum speed of the object = 0.62 m/s