Question

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes 2 cm, $1m{s}^{-1}$ and $10m{s}^{-2}$ at a certain instant. Find the amplitude and the time period of the motion.

Answer 1

Step 1: Given that:

At a certain instant;

The magnitude of the position of the particle executing SHM(x) = $2cm$ = $0.02m$

The magnitude of the velocity of the particle executing SHM(v) = $1m{s}^{-1}$

The magnitude of the acceleration of the particle executing SHM(a)= $10m{s}^{-2}$

Step 2: Formula used:

The acceleration of a particle executing SHM is given by;

$a={\text{ω}}^{2}x$

$\text{ω}=\sqrt{\frac{a}{x}}$

Where ω is the angular velocity of the particle.

$\text{ω}=\frac{2\pi }{T}$

$T=\frac{2\pi }{\text{ω}}$

The velocity of the particle executing SHM is given as;

$v=\text{ω}\sqrt{(A^2-x^2)}$

$\Rightarrow \text{ω}\sqrt{(A^2-x^2)}=v$

$\Rightarrow {\omega }^2(A^2-x^2)=v^2$

$\Rightarrow (A^2-x^2)=\frac{v^2}{{\omega }^2}$

$\Rightarrow A^2=\frac{v^2}{{\omega }^2}+x^2$

$\Rightarrow A=\sqrt{(\frac{v^2}{{\omega }^2}+x^2)}$

Where A is the amplitude of the particle.

Step 3: Calculation of time period of the particle:

$\text{ω}=\sqrt{\frac{10m{s}^{-2}}{0.02}}$

$\text{ω}=\sqrt{\frac{1000}{2}}$

$\text{ω}=\sqrt{500}$

$\text{ω}=10\sqrt{5}$

Now,

$T=\frac{2\pi }{10\sqrt{5}}$

$T=\frac{2\times 3.14}{10\times 2.236}$

$T=\frac{6.28}{22.36}$

$T=0.28sec$

Step 4: Calculation of amplitude of the particle:

$A=\sqrt{(\frac{v^2}{{\omega }^2}+x^2)}$

Putting the values in above we, get;

$A=\sqrt{(\frac{{\left(1m{s}^{-1}\right)}^2}{{\left(10\sqrt{5}\right)}^2}+{\left(0.02m\right)}^2)}$

$A=\sqrt{(\frac{1}{500}+0.0004)}$

$A=\sqrt{0.0020-0.0004}$

$A=\sqrt{0.0016}$

$A=0.04m$

Thus,

The time period of the particle = 0.28sec

The amplitude of the particle = 0.04m = 4cm