Question

In the device shown in the figure, the block of mass $20kg$ is displaced down slightly and released, it starts oscillating. Pulleys are smooth and massless. Strings and springs are also massless. Find the time period of oscillation (${{K}_{1}}=4N/m,{{K}_{2}}=4N/m$)?

Answer 1

Hint: Before solving this question one should know that simple harmonic motion is only observed when acceleration is proportional to displacement with degree one. In SHM, we can replace acceleration with equation ${{\omega }^{2}}x$which will help us in finding the time period of the motion of the object.

Formula used:
Tension $T=Kx$(equal to spring force) and $\omega =2\pi f$.

Complete step by step answer:
Let’s consider the elongation in spring with spring constant ${{K}_{1}}$ as${{x}_{1}}$ and ${{K}_{2}}$ as ${{x}_{2}}$ when the mass is being displaced by $x$ units.Now, since the string is the same, displacement of mass will be equal to the net displacement of string. Spring elongation in ${{K}_{1}}$ will be equal to the elongation in the string attached to it. But here comes a problem: the string is rolled over the pulley so the displacement in the string will be divided equally which is $x/2$.

Talking about the spring with constant ${{K}_{2}}$ since we have assumed its elongation to be ${{x}_{2}}$ the elongation in the string directly attached to it will also be the same, and because of spring with constant ${{K}_{1}}$the pulley attached to that string will create a displacement of ${{x}_{1}}/2$which have to be equal to ${{x}_{2}}$ because of string property. Since the whole system is in equilibrium, we can equate the tension as well.

From the above discussion we get the following result: ${{x}_{2}}={{x}_{1}}/2$
The above relation will be helpful in finding net displacement in terms of string elongation of strings attached to springs and rolled over pulleys.
Now let’s equate the tension to get another relation $2{{K}_{1}}{{x}_{1}}={{K}_{2}}{{x}_{2}}$putting the values of spring constant as provided in question we will get
$2{{x}_{1}}={{x}_{2}}$

Now let’s write total displacement
$x={{x}_{1}}+2{{x}_{2}}$
Putting the value of ${{x}_{2}}$ as obtained in the relation we will get
$x=5{{x}_{1}}$
Now restoring force, occurring on the mass will be ${{K}_{1}}{{x}_{1}}=ma$ where a is the acceleration by definition of SHM we can replace acceleration with ${{\omega }^{2}}x$ (see the hint).
${{K}_{1}}{{x}_{1}}=m{{\omega }^{2}}(5{{x}_{1}})$
On solving and putting the value of spring constant $\omega =1/5$
We know $\omega =2\pi f$ where $f$ is frequency and relation between frequency and time period is $f=1/T$. On solving we will get $T=10\pi $ seconds as the answer.

Hence, the time period of oscillation is $10\pi $.

Note: Such a question should be done by visualizing the situation since that will help you in understanding the concept of the spring system. Try to equate the tension and find the net displacement, use both to reach your desired answer.