Question

A $150\text{g}$ block oscillates in SHM on the end of a spring with $k=1500\text{N/m}$, according to the relation $x=x_0\cos (\omega t+\varphi ).$ How long does the block take to move from the position $+0.800x_0$ to $+0.600x_0$ directly ?

• A. $2\text{ms}$
• B. $0.3\text{ms}$
• C. $4\text{ms}$
• D. $3\text{ms}$

Answer 1

The correct option is D $3\text{ms}$

Given that
mass of block $\left(m\right)=150\text{g}\text{or}0.15\text{kg}$
spring constant $\left(k\right)=1500\text{N/m}$.
Equation of SHM, $x=x_0\cos (\omega t+\varphi )$
We know that,
$\omega =\sqrt{\frac{k}{m}}$
From the given data,
$\omega =\sqrt{\frac{1500}{0.15}}=100\text{rad/s}$

Equation of SHM is $x=x_0\cos (\omega t+\varphi )$
Let $t_1$ be the time at which $x=+0.8x_0$.
Then, $0.8x_0=x_0\cos (\omega t+\varphi )$
$\Rightarrow \omega t_1+\varphi ={\cos }^{-1}\left(0.8\right)=0.6435\text{rad}.....\left(1\right)$

Similarly, when it at $x=+0.6x_0$ at time $t_2$,
$\omega t_2+\varphi ={\cos }^{-1}\left(0.6\right)=0.9272\text{rad}....\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$|\omega (t_2-t_1)|=0.2837\text{rad}$
$\Rightarrow t_2-t_1=\frac{0.2837}{\omega }=\frac{0.2837}{100}\approx 3\text{ms}$