First slide

Applications of SHM

Question

A constant force produces maximum velocity v on the block connected to the spring of force constant k as shown in the figure. When the force constant of spring becomes 4k,the maximum velocity of the block is (block is at rest when spring is relaxed):

Difficult

A

v/4
B

2v
C

v/2
D

v

Solution

By work energy theorem;
$\begin{array}{l} F x_{1} - \frac{1}{2} k x_{1}^{2} = \frac{1}{2} m v^{2} . . . . . . . (1) \\ and F x_{2} - \frac{1}{2} k^{'} x_{2}^{2} = \frac{1}{2} m v^{' 2} . . . . . . . . . (2) \end{array}$
where x₁,x₂are initial and final extensions and v,t' are initial
and final velocities.
In both cases, force applied is same, and velocity becomes maximum when F=kx. (after which the mass will decelerate)
$\begin{array}{l} ∴ F = k x_{1} = (4 k) x_{2} \\ \Rightarrow x_{2} = \frac{x_{1}}{4} \end{array}$
Substituting in (2):
$\begin{array}{l} F (\frac{x_{1}}{4}) - \frac{1}{2} (4 K) {(\frac{x_{1}}{4})}^{2} = \frac{1}{2} m v^{' 2} \\ \Rightarrow \frac{1}{4} [F x_{1} - \frac{1}{2} k x_{1}^{2}] = \frac{1}{2} m v^{' 2} . . . . . . . (3) \end{array}$
Dividing (3)/(1), we get:
$\frac{1}{4} = \frac{v^{' 2}}{v^{2}} \Rightarrow v^{'} = \frac{v}{2}$
Hence, (c).

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