Question

A particle executes simple harmotic motion about the point x=0. At time t=0, it has displacement x=2cm and zero uelocity. If the frequency of motion is 0.25 s_(-1), find (a) the period, (b) angular frequency, ( c) the amplitude, (d) maximum speed, (e) the displacement from the mean position at t=3 sand (f) the velocity at t=3 s.

Answer 1

Given :

The point about which the particle is executing S.H.M. is :

x = 0

The displacement and velocity of the particle at t = 0 :

x = 2 cm and v=0

The frequency of given S.H.M . is :

f = 0.25 $s^-$

To Find :

a). time period

b).angular frequency

c). the amplitude

d).maximum speed

e). the displacement from the mean position at t = 3 s

f). velocity at t = 3 s

Solution :

a).We know that time period   $T = \frac{1}{f}$

So, $T = \frac{1}{0.25}s$

⇒T =  4 sec

b).We know that formula for angular speed ω is :

$\omega = \frac{2\pi}{T}$

So, $\omega = \frac{2\pi}{4}$

Or,$\omega =\frac{\pi}{2} \frac{rad}{s} = 1.57 \frac{rad}{s}$

c). Since we know that the amplitude is the maximum displacement from the mean position , and here mean position is x = 0 , and since the particle is having 0 velocity at x = 2 cm , so its maximum displacement from the mean position is 2 cm

So, amplitude A = 2 - 0 cm = 2 cm

d).Maximum speed of the particle in S.H.M. is given as :

$V_{max}= A \omega$

So, $V_{max} =2 \times \frac{\pi}{2}$

 $V_{max} = 3.14 \frac{cm}{s}$

e). Since , at t=0 particle starts from extreme position , so its equation is given as :

$x = A cos\omega t$

  $= A cos (\frac{\pi}{2}t)$

So,at t = 3 s , x will be :

$x = A cos (\frac{\pi}{2}\times 3)$

   $= A \times 0$

   = 0

So, at t= 3 s , displacement from the mean position of the particle is 0.

f). Since from previous answer we know that at t = 3 sec , x = 0 i.e. particle is at mean position and we also know that at mean position velocity of the particle is maximum , so :

$V_{t=3s} = V_{max} = 3.14 \frac{cm}{s}$

Answer 2

(a) The period is 4 seconds

(b) The angular frequency is π/2 rad/s

(c) The amplitude is 2 cm

(d) The maximum speed is π cm/s

(e) The displacement from the mean position at t = 3 s is 0

(f) The velocity at t = 3 s is π cm/s

Explanation:

(a) The time period is given by the formula:

T = 1/f

Where,

f = Frequency = 0.25 s⁻¹

On substituting the values, we get,

T = 1/0.25

∴ T = 4 seconds

(b) The angular frequency is given by the formula:

ω = 2π/T

On substituting the values, we get,

ω = 2π/4

∴ ω = π/2 rad/s

(c) The amplitude is given as maximum displacement from mean position.

A = 2 - 0

∴ A = 2 cm

(d) The maximum speed is given by the formula:

v = Aω

On substituting the values, we get,

A = 2 × π/2

∴ v = π cm/s

(e) The position of the particle is given by the equation:

x = A cos(ωt)

Where,

t = 3 seconds

On substituting the values, we get,

x = A cos(π/2 × 3) = A cos(3π/2)

∵ 3π/2 = 0

x = A × 0

∴ x = 0

(f) The velocity of the particle at 3 seconds is at zero position. Thus, the maximum velocity is:

∴ Vmax = π cm/s