Question

A block of mass $m$, attached to a spring of spring constant $k$, oscillates on a smooth horizontal table. The other end of the spring is fixed to a wall. The block has a speed $v,$ when the spring is at its natural length. How far will the block move on the table before coming to an instantaneous rest?

• A. $x=2v\sqrt{\frac{m}{k}}$
• B. $x=\frac{1}{v}\sqrt{\frac{m}{k}}$
• C. $x=v\sqrt{\frac{m}{k}}$
• D. $x=3\sqrt{\frac{mv}{k}}$

Answer 1

The correct option is C $x=v\sqrt{\frac{m}{k}}$

In SHM, speed of the block is maximum at mean position and speed is zero at extreme position.
For given case, particle is at mean position, when spring is in its natural length.
Maximum velocity in SHM $v=A\omega$
Angular frequency of the oscillation is
$\Rightarrow \omega =\sqrt{\frac{k}{m}}$

$\Rightarrow A=\frac{v}{\omega }$
$\Rightarrow A=v\sqrt{\frac{m}{k}}$
When mass is at instantaneous rest, position of particle from mean position is $A=v\sqrt{\frac{m}{k}}$
Hence option 'c' is correct


OR

From conservation of mechanical energy,
$K_i+U_i=K_f+U_f$

Initial kinetic energy is $K_i=\frac{1}{2}m{v}^2$
As spring is in its natural length, initial potential energy $U_i=0$
For mass to be at rest, final kinetic energy is $K_f=0$
As spring stretches by $x$, final potential energy $U_f=\frac{1}{2}k{x}^2$

$k{x}^2=m{v}^2$
$x=v\sqrt{\frac{m}{k}}$