Question

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 7.5 cm and released. The ball makes 31 oscillations in 15 s seconds.1) What is its the mass of the ball?Express your answer using two significant figures.m = _____ g2) What is its maximum speed?Express your answer using two significant figures.= _____ cm/s

Answer 1

Explanation:

It is given that,

Spring constant of the spring, k = 15 N/m

Amplitude of the oscillation, A = 7.5 cm = 0.075 m

Number of oscillations, N = 31

Time, t = 15 s

(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :

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

Total number of oscillation per unit time is called frequency of oscillation. Here, $f=\dfrac{31}{15}=2.06\ Hz$

$m=\dfrac{k}{4\pi^2f^2}$

$m=\dfrac{15}{4\pi^2\times 2.06^2}$

m = 0.0895 kg

or

m = 89 g

(b) The maximum speed of the ball that is given by :

$v_{max}=A\times \omega$

$v_{max}=A\times 2\pi f$

$v_{max}=0.075\times 2\pi \times 2.06$

$v_{max}=0.970\ m/s$

$v_{max}=97\ cm/s$

Hence, this is the required solution.

Answer 2

The mass of ball be "89 g" and maximum speed be "97 cm/s".

Mass and Speed:

Given values are:

Amplitude of oscillation, A = 7.5 cm or, 0.075 m

Number of oscillation, N = 31

Spring constant, k = 15 N/m

Time, t = 15 s

then,

Frequency, f = $\frac{31}{15}$ = 2.06 Hz

We know the relation,

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

or,

→ $m = \frac{k}{4 \pi^2 f^2}$

By substituting the values, we get

      $= \frac{15}{4 \pi^2\times (2.06)^2}$

      $= 0.0895$ kg or, 89 g

and,

Ball's maximum speed will be:

→ $v_{max}$ = A × ω

or,

          = A × 2πf  

By substituting the values,

          = $0.075\times 2 \pi\times 2.06$

          = $0.970$ m/s or, $97$ cm/s

Thus the above answers are correct.

Find out more information about speed here:

https://brainly.com/question/25959744