First slide

Applications of SHM

Question

Block A of mass 2kg moving with 10 m/s strikes a spring of constant $π^{2} N / m$ attached to another 2kg block B at rest kept on a smooth floor. The time for which rear moving block A remain in contact with spring will be

Difficult

A

$\sqrt{2} s$
B

$\frac{1}{\sqrt{2}} s$
C

$1 s$
D

$\frac{1}{2} s$

Solution

According to the statement
$m_{1} = m_{2} = 2 kg$
We have a relation for the time period of two spring mass system
$t = (2 π) \sqrt{(\frac{μ}{k})} ..... (1)$
Where k=spring constant= $π^{2} N / m$
$μ =$ reduced mass= $\frac{m_{1} m_{2}}{m_{1} + m_{2}}$
The time period of both oscillations are same
$(2 π) \sqrt{\frac{m_{1}}{k_{1}}} = π \sqrt{\frac{m_{1}}{k_{2}}} .... (2)$
Now calculations for $μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}$
Substituting the values
$μ = \frac{2 \times 2}{2 + 2} = \frac{4}{4} = 1$
For $μ = 1$
Substituting the value of $' μ' &' k'$ in (1)
$T = (2 π) \sqrt{\frac{1}{π^{2}}}$
$T = 2 s$
But it takes $\frac{T}{2}$ time for first block to detach, implies time $\frac{T}{2} = 0.5 \times 2 = 1 s$
The time for which rear moving block remain in contact with spring will be 1 second.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App