Share

## At an outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 29.8981 cm on a spring with a spring constant o

Question

At an outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 29.8981 cm on a spring with a spring constant of 19.1461 N/m. The mass of the bananas is 43.288 kg. What is the maximum speed of the bananas

in progress
0

Physics
3 years
2021-07-27T08:28:15+00:00
2021-07-27T08:28:15+00:00 1 Answers
12 views
0
## Answers ( )

Answer:

Explanation:

Given that,

The amplitude of oscillation is

A = 29.8981cm = 0.298981m

Spring constant k = 19.1461N/m

Mass of banana hung m = 43.288kg

Maximum speed v?

The maximum speed can be determined by using the formula

v = w•A

Where

w = angular frequency

A = amplitude

So we need to get the angular frequency and it can be determined by using the formula

w = √ (k/m)

w = √(19.1461/43.288)

w = √0.4423

w = 0.6651rad/sec

Then, maximum speed

v = w•A

v = 0.6651 × 0.298981

v = 0.19884m/s

v ≈ 0.2m/s

The maximum speed of oscillation is 0.2m/s