Request a call back

Join NOW to get access to exclusive study material for best results

CBSE Class 11-science Answered

A small body of mass 0.10kg is undergoing simple harmonic motion of simplitude 1.0 meter and period of 0.20 sec the maximum value of the force acting on it
Asked by devvratagrahari | 02 Dec, 2019, 06:44: AM
answered-by-expert Expert Answer
The value of maximum force is, Fmax = mamax ... (1)
We know, amax = Aω2 
Now, T = 2∏/ω
→ω = 2∏/T 
 
Substituting in (1), we get, 
Fmax = m A(2∏/T)2 
Thus, 
Fmax = 0.1×1 × [(2 × 3.14)/0.2]2 = 98.596 N
 
 
Answered by Shiwani Sawant | 02 Dec, 2019, 03:22: PM
CBSE 11-science - Physics
Asked by sakshiiiyadav9336 | 02 Feb, 2024, 09:37: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by snehasahu824 | 03 Mar, 2020, 10:06: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by Topperlearning User | 15 May, 2015, 10:54: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
Get Latest Study Material for Academic year 24-25 Click here
×