Question

A 25.0 kg mass is traveling to the right with a speed of 2.80 m/s on a smooth horizontal surface when it collides with and sticks to a second 25.0 kg mass that is initially at rest but is attached to one end of a light, horizontal spring with force constant 170.0 N/m. The other end of the spring is fixed to a wall to the right of the second mass.

1) Find the frequency of the subsequent oscillations.2) Find the amplitude of the subsequent oscillations.3) Find the period of the subsequent oscillations.4) How long does it take the system to return the first time to the position it had immediately after the collision?

Answer 1

The frequency of the subsequent oscillations is 0.487 Hz. The amplitude of the subsequent oscillations is 0.411 m. The period of the subsequent oscillations is 2.05 seconds. It takes approximately 2.04 seconds for the system to return to the position it had immediately after the collision.

Step-by-step explanation:

1) Frequency of the subsequent oscillations:

The frequency of the subsequent oscillations can be determined using the formula:

f = (1/(2 ×
`\pi`)) × √(k/m)

Where f is the frequency, k is the force constant of the spring, and m is the mass of the system.

Plugging in the values gives:

f = (1/(2 ×
`\pi`)) × √(170 N/m / 50 kg) = 0.487 Hz

Therefore, the frequency of the subsequent oscillations is approximately 0.487 Hz.

2) Amplitude of the subsequent oscillations:

The amplitude of the subsequent oscillations can be determined using the formula:

A = (m × v)/(k)

Where A is the amplitude, m is the mass of the system, v is the initial velocity of the mass, and k is the force constant of the spring.

Plugging in the values gives:

A = (25 kg × 2.80 m/s) / (170 N/m) = 0.411 m

Therefore, the amplitude of the subsequent oscillations is approximately 0.411 m.

3) Period of the subsequent oscillations:

The period of the subsequent oscillations can be determined using the formula:

T = 1/f

Where T is the period and f is the frequency.

Plugging in the value of the frequency (0.487 Hz) gives:

T = 1 / 0.487 = 2.05 s

Therefore, the period of the subsequent oscillations is approximately 2.05 seconds.

4) Time taken to return to the initial position:

The time taken to return to the initial position can be determined using the formula for the period:

T = 2 ×
`\pi` × √(m/k)

Where T is the period, m is the mass of the system, and k is the force constant of the spring.

Plugging in the values gives:

2 ×
`\pi` × √(50 kg / 170 N/m) = 2.04 s

Therefore, it takes approximately 2.04 seconds for the system to return to the position it had immediately after the collision.

Answer 2

Step-by-step explanation:

1 ) angular frequency ω = √ ( k / m )

=√ ( 170 / 50 )

= 1.844 rad /s

2πn = 1.844 where n is frequency of oscillation

n = 1.844 / (2 x 3.14 )

= .294 per sec

= .294 x 60 = 18 approx. per minute .

Velocity just after collision of composite mass ( using law of conservation of momentum )

= 25 x 2.8 / 50

v = 1.4 m/s

If new amplitude be A

1/2 k A² = 1/2 m v²

m = 25 + 25 = 50 kg

170 x A² = 50 x 1.4²

A = 0.76 m

3 ) period of oscillation = 1 /n

= 1 / .294

= 3.4 s

4 ) It will take complete one period of oscillation ie 3.4 s to come to its original position.