An oscillating block of mass 250 g takes 0.15 sec to move between the endpoints of motion, which are 40cm apart.

  1. What is the frequency of the motion?
  2. What is the amplitude of the motions?
  3. What is the forces constant of the spring?

Answers

This answer is not selected as best answer. This answer may not be sufficient for exam.

Your limit has been exceed. We have implemented this system because, We got difficulty on managing our servers. Please donate some amount to remove this limit.

Quota: 0 / 30

Donate

Solution:

User Loaded Image | CSIT Guide

Here is given, block of mass (m) = 250gm = 250 x 10-3 kg

Time (t) = 0.15 sec

Displacement (x) = 40cm = 40 x 10-2m

Frequency (f) = ?

Force constant (k) = ?

We know,

a.  f  = 1/t = 1/0.15 = 6.67 rev sec-1

b. Amplitude (xm) =

\(\frac{40 \times 10^{-2}}{2}\) = 20 x 10-2 m

c. Again we know that

\(2\pi f = \sqrt{\frac{k}{m}}\)

\(4\pi^2f^2 = \frac{k}{m}\)

\(k = 4\pi^2f^2 m\)

\(k = 4\pi^{2}(6.67)^2 \times 250 \times 10^{-3}\)

k = 439.08 Nm-1

Hence, required frequency, amplitude and force constant are 6.67 rev s-1, 20 x 10-2m and 439.08 Nm-1

If you found any type of error on the answer then please mention on the comment or submit your new answer.
Leave your Answer:

Click here to submit your answer.

Discussion
0 Comments
  Loading . . .