Question

A block with a mass of $1\text{kg}$ is connected to a massless spring with a force constant of $100\text{N/m}$ and is placed on a smooth horizontal surface. At $t=0$, a constant force of $F=10\text{N}$ is applied on the block, when the spring is in its natural length. The speed of the block, when it is displaced by $8\text{cm}$ from its mean position will be

• A. $80\text{cm/s}$
• B. $16\text{cm/s}$
• C. $60\text{cm/s}$
• D. $25\text{cm/s}$

Answer 1

The correct option is C $60\text{cm/s}$

Due to force $F$, the block will starts moving in the direction of force and thus a restoring force is developed in the spring. Therefore, the motion of the block will be simple harmonic motion.

Considering the natural length position of the spring as the refrence point (origin).
Let $x_{mean}$ be the mean position of the SHM from the origin.
$\Rightarrow F=-k{x}_{mean}$
$\Rightarrow x_{mean}=\mid \frac{F}{k}\mid =\frac{10}{100}=0.1\text{m}=10\text{cm}$

Let the maximum displacement of the block be $x_{max}$ from the origin.
Applying work energy theorem :
$W=\Delta K.E$
$W_{spring}+W_F=K.E_f-K.E_i$
$-\frac{1}{2}K{x}_{max}^2+F{x}_{max}=0-0$
$(\because K.E_f=0$ at extreme position$)$
$-\frac{1}{2}\times 100x_{max}^2+10x_{max}=0$
$\Rightarrow x_{max}=\frac{2}{10}=0.2\text{m}=20\text{cm}$

Therefore, amplitude of the SHM $\left(A\right)$$=x_{max}-x_{mean}=20\text{cm}-10\text{cm}$
$A=10\text{cm}$

Here, $k$ is the force constant
The angular frequency $\left(\omega \right)$ is given by :
$\omega =\sqrt{\frac{k}{m}}\Rightarrow \omega =\sqrt{\frac{100}{1}}=\omega =10\text{rad/s}$

The expression for the speed of the block, when it is displaced by $x=8\text{cm}$ from its mean position is :
$v=\omega \sqrt{A^2-x^2}$
$=10\sqrt{(10{)}^2-(8{)}^2}=60\text{cm/s}$
Thus, option (c) is the correct answer.