Question

H Oscllatons a and Due Dates >Hw Osoillations and Waves 1 Resources Hint A 0.250-kg particle undergoes simple harmonic motion along the horizontal -axis between the points z-0.225 m and 0.439 m. The period of oscillation is 0.641 s. Find the frequency, f, the equilibrium position, , the amplitude, A, the maximum speed, , the maximum magnitude of acceleration, mas, the force constant, k, and the total mechanical energy, E Hz m/s A= m m/s N/ Gmas E ana ayp a hea sbouta P3

Answer 1

Given:-

• Mass of the particle :
  m = 0.259 kg
  
• Position                   :
  01 = -0.225 m
  
• Position                   :
  r2 = 0.439 m
  
• Period of oscillation   :
  T = 0.641 s
  

Part (1) Frequency :

Frequency is reciprocal of period of oscillation,

$f = \frac{1}{T}$

f = 0.641

f= 1.56 Hz

This is the frequency of oscillations.

Part (2) Equilibrium position :

Equilibrium position is given by,

11+12 Veq =

-0.225 +0.439 Ieq =

Xeq = 0.107 m

This is the equilibrium position.

Part (3) Amplitude :

Amplitude is given by,

12-21 A =

0.439 +0.225 A =

A=0.332 m

Part (4) Maximum speed :

Maximum speed is given by,

Umar = -WA

Where, is angular speed of the SHM.

f = لي

Umar = -27A

Umar = -2 x 3.14 x 1.56 x 0.332

Vmax = -3.25 m/s

This is the maximum speed of motion of particle. Negative sign indicates the direction of the speed, thus magnitude of maximum speed will be 3.25 m/s.

Part (5) Acceleration :

Maximum acceleration is given by,

amar = -w24

Amar = -(2) A

Amar = - (2 x 3.14 x 1.56)2 x 0.332

amar = -31.9

amax = 31.9 m/s2

This is magnitude of the maximum acceleration.

Part (6) Force constant :

Force constant k is given by,

է: = mar

k=mx (275)

k=0.259 x (2 x 3.14 x 1.56)

k= 24.9 N/m

Part (7) Total mechanical energy :

Total mechanical energy is given by,

Exot = (4) k4= Etot =

Etot = 2 x 24.9 (0.332)

Etot = 1.4J