1. ⇒ (MHT CET 2023 12th May Morning Shift )

Four massless springs whose force constants are $2 K, 2 K, K$ and $2 K$ respectively are attached to a mass $M$ kept on a frictionless plane as shown in figure, If mass $M$ is displaced in horizontal direction then frequency of oscillating system is

MHT CET 2023 12th May Morning Shift Physics - Simple Harmonic Motion Question 3 English

A. $\frac{1}{2 π} \sqrt{\frac{K}{4 M}}$

B. $\frac{1}{2 π} \sqrt{\frac{4 K}{M}}$

C. $\frac{1}{2 π} \sqrt{\frac{K}{7 M}}$

D. $\frac{1}{2 π} \sqrt{\frac{7 K}{M}}$

Correct Option is (B)

On the right hand side of the block, springs are connected in parallel

$∴$ Their effective spring constant is given by

$\begin{aligned} K_{1} = K + 2 K \\ K_{1} = 3 K \end{aligned}$

On the left hand side of the block, springs are connected in series.

$∴$ Their effective spring constant is given by,

$\begin{aligned} \frac{1}{K_{2}} = \frac{1}{2 K} + \frac{1}{2 K} \\ ∴ & K_{2} = K \end{aligned}$

$∴$ Effective spring constant of the system is given by,

$\begin{aligned} K_{E} = 3 K + K = 4 K \\ ∴ ω = \sqrt{\frac{K_{E}}{M}} = \sqrt{\frac{4 K}{M}} \\ ∴ f = \frac{ω}{2 π} = \frac{1}{2 π} \sqrt{\frac{4 K}{M}} \end{aligned}$

2. ⇒ (MHT CET 2023 11th May Evening Shift )

The upper end of the spring is fixed and a mass ' $m$ ' is attached to its lower end. When mass is slightly pulled down and released, it oscillates with time period 3 second. If mass ' $m$ ' is increased by $1 kg$ , the time period becomes 5 second. The value of ' $m$ ' is (mass of spring is negligible)

A. $\frac{3}{8} kg$

B. $\frac{5}{9} kg$

C. $\frac{8}{13} kg$

D. $\frac{9}{16} kg$

Correct Option is (C)

The formula for the time period of a spring mass system is $T = 2 π \sqrt{\frac{m}{k}}$

For mass $m + 1, T^{'} = 2 π \sqrt{\frac{m + 1}{k}}$

Taking the ratio,

$\begin{aligned} \frac{T}{T^{'}} & = \frac{2 π \sqrt{\frac{m}{k}}}{2 π \sqrt{\frac{m + 1}{k}}} \\ \frac{T}{T^{'}} & = \sqrt{\frac{m}{m + 1}} \\ \frac{3}{5} & = \sqrt{\frac{m}{m + 1}} \\ \frac{9}{25} & = \frac{m}{m + 1} \\ ∴ 9 m & + 9 = 25 m \\ ∴ 16 m & = 9 \\ ∴ m & = \frac{9}{16} kg \end{aligned}$

3. ⇒ (MHT CET 2023 9th May Evening Shift )

A spring has a certain mass suspended from it and its period for vertical oscillations is ' $T_{1}$ '. The spring is now cut in to two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now ' $T_{2}$ '. The ratio $T_{1} / T_{2}$ is

A. 2

B. $\sqrt{2}$

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

D. $\frac{1}{2}$

Correct Option is (B)

$T_{1} = 2 π \sqrt{\frac{m}{k}} .$

Also, spring constant $(k) \propto \frac{1}{Length (l)}$

When the spring is half in length, then $k$ becomes twice.

$\begin{array}{ll} ∴ & T_{2} = 2 π \sqrt{\frac{m}{2 k}} \\ ∴ & \frac{T_{1}}{T_{2}} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \end{array}$

4. ⇒ (MHT CET 2021 20th September Evening Shift )

A body performs S.H.M. under the action of force ' $F_{1}$ ' with period ' $T_{1}$ ' second. If the force is changed to ' $F_{2}$ ' it performs S.H.M. with period ' $T_{2}$ ' second. If both forces ' $F_{1}$ ' and ' $F_{2}$ ' act simultaneously in the same direction on the body, the period in second will be

A. $\frac{T_{1} + T_{2}}{T_{1} T_{2}}$

B. $\frac{T_{1}^{2} + T_{2}^{2}}{T_{1} T_{2}}$

C. $\frac{T_{1} T_{2}}{\sqrt{T_{1}^{2} + T_{2}^{2}}}$

D. $\frac{T_{1} T_{2}}{T_{1} + T_{2}}$

Correct Option is (C)

$\begin{aligned} F_{1} = K_{1} x F_{2} = K_{2} x F = (K_{1} + K_{2}) x \\ ∴ T_{1} = 2 π \sqrt{\frac{m}{K}} T_{2} = 2 π \sqrt{\frac{m}{K}} \\ ∴ T_{1}^{2} = 4 π^{2} \frac{m}{K_{1}} T_{2}^{2} = 4 π^{2} \frac{m}{K_{2}} \\ T^{2} = 4 π^{2} \frac{m}{K_{1} + K_{2}} \\ ∴ \frac{1}{T^{2}} = \frac{K_{1} + K_{2}}{4 π^{2} m} = \frac{K_{1}}{4 π^{2} m} + \frac{K_{2}}{4 π^{2} m} \\ = \frac{1}{T_{1}^{2}} + \frac{1}{T_{2}^{2}} = \frac{T_{1}^{2} + T_{2}^{2}}{T_{1}^{2} T_{2}^{2}} \\ ∴ T^{2} = \frac{T_{1}^{2} T_{2}^{2}}{T_{1}^{2} + T_{2}^{2}} \\ ∴ T = \frac{T_{1} T_{2}}{\sqrt{T_{1}^{2} + T_{2}^{2}}} \end{aligned}$

5. ⇒ (MHT CET 2021 20th September Morning Shift )

A mass ' $m_{1}$ ' is suspended from a spring of negligible mass. A spring is pulled slightly in downward direction and released, mass performs S.H.M. of period ' $T_{1}$ '. If the mass is increased by ' $m_{2}$ ', the time period becomes ' $T_{2}$ '. The ratio $\frac{m_{2}}{m_{1}}$ is

A. $\frac{T_{1}^{2} + T_{2}^{2}}{T_{1}^{2}}$

B. $\frac{T_{1} - T_{2}}{T_{1}}$

C. $\frac{T_{2}^{2} - T_{1}^{2}}{T_{1}^{2}}$

D. $\frac{T_{1}^{2} - T_{2}^{2}}{T_{1}^{2}}$

Correct Option is (C)

$\begin{aligned} T_{1} = 2 π \sqrt{\frac{m_{1}}{k}}, T_{2} = 2 π \sqrt{\frac{m_{1} + m_{2}}{k}} \\ ∴ \frac{T_{2}}{T_{1}} = \sqrt{\frac{m_{1} + m_{2}}{m_{1}}} \\ ∴ \frac{T_{2}^{2}}{T_{1}^{2}} = \frac{m_{1} + m_{2}}{m_{1}} \\ ∴ \frac{T_{2}^{2} - T_{1}^{2}}{T_{1}^{2}} = \frac{m_{2}}{m_{1}} \end{aligned}$

⇒Oscillations