A $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{kg}$block sliding on a horizontal frictionless surface is attached to a horizontal spring with role="math" localid="1657267407759" $\mathbf{k}\mathbf{=}\mathbf{480}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{m}$. Let $\mathrm{x}$be the displacement of the block from the position at which the spring is unstretched $\mathbf{t}\mathbf{=}\mathbf{0}$. At the block passes through $\mathbf{x}\mathbf{=}\mathbf{0}$with a speed of $\mathbf{5}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$in the positive $\mathbf{x}$direction. What are the (a) frequency and (b) amplitude of the block’s motion? (c) Write an expression for$\mathbf{x}$as a function of time.

• The mass of the block is, $\mathrm{M}=1.2\mathrm{kg}$.
• The displacement of the block from equilibrium position is $\mathrm{x}$.
• The speed of the block at $\mathrm{t}=0$when $\mathrm{x}=0$is,${\mathrm{v}}_{\mathrm{m}}=5.2\mathrm{m}/\mathrm{s}$.
• The spring constant is,$\mathrm{k}=480\mathrm{N}/\mathrm{m}$.

We can find the frequency of the block’s motion from angular frequency. Then we can find the amplitude of the block’s motion using the formula for the maximum velocity of SHM. Next, we can find the expression for x as a function of time for the block’s motion from the expression for the displacement of SHM.

Formulae:

The maximum speed of SHM, $v_m=\omega x_m$

(i)

The angular frequency of oscillation in S.H.M, $\omega =\sqrt{\frac{k}{m}}$ (ii)

The frequency of the block’s motion, $f=\frac{\omega }{2\mathrm{\pi }}$ (iii)

From equation (ii) and the given values, we get the angular frequency of the motion as:

$$\begin{gathered} \omega =\sqrt{\frac{480\mathrm{N}/\mathrm{m}}{1.2\mathrm{kg}}} \\ =20\mathrm{rad}/\mathrm{s} \end{gathered}$$

Using equation (iii), the frequency of the oscillation is given as:

$\begin{array}{l}f=\frac{20\mathrm{rad}/\mathrm{s}}{2\times 3.142}\\ =3.18\\ \approx 3.2\mathrm{Hz}\end{array}$

Therefore, the frequency of the block’s motion is $3.2\mathrm{Hz}$.

Using equation (i), the amplitude of the oscillation is given as:

$$\begin{gathered} x_m=\frac{v_m}{\omega } \\ =\frac{5.2}{20} \\ =0.26\mathrm{m} \end{gathered}$$

Therefore, the amplitude of the block’s motion is$0.26\mathrm{m}$.

The general expression of displacement of the SHM is given as:

$x=x_m\cos (\omega t+\varphi )$

The displacement of the block’s motion is,

$x=0.26\mathrm{m}\cos (20t+\varphi )$

We have given thatlocalid="1657268704017" $x=0\mathrm{at}\mathrm{t}=0$.Then,

localid="1657269100980" $\begin{array}{ll}0=(0.26m)\cos \left(20\right(0& m)+\varphi )\\ \mathrm{co}s\varphi =0& \\ \varphi =\pm \frac{\pi }{2}& \end{array}$

The velocity of the SHM is,

$v=-v_m\sin (\omega t+\varphi )$

The velocity of the block’s motion is,

$v=(-5.2\mathrm{m})\sin (20t+\varphi )$

At,$\varphi =\frac{\pi }{2}$,localid="1657269042132" $v$becomes zero.

Hence, the accepted phase angle of the motion is:

$\varphi =-\frac{\pi }{2}$

Therefore, the displacement of the block’s motion is, localid="1657269307179" $x=(0.26)cos\left(20t-\frac{\pi }{2}\right)\mathrm{m}.$