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PHXI14:OSCILLATIONS

364398 A uniform cylinder of mass $m = 2 k g$ and radius $R = 0.50$ $m$ is in equilibrium on an inclined plane by the action of a light spring of stiffness $K = 300 N / m$ , gravity and reaction force acting on it. If the angle of inclination of the plane is $ϕ = 30^{\circ}$ , find the angular frequency of small oscillation of the cylinder.
supporting img

1

20 r a d / s

2

55 r a d / s

3

29 r a d / s

4

32 r a d / s

Explanation:

Here point of contact $P$ , will be instantaneous centre of rotation. Let the cylinder rotates a small angle $θ$ beyond its equilibrium position about an axis passing through $P$ . Here the increase in length of spring will be $2 R θ$ . Thus the increased spring force provides a restoring torque about $P$ .
supporting img
Now writing torque equation about $P$ as:
$Δ τ_{p} = I_{p} α$
$(K \cdot 2 R θ) \cdot 2 R = [\frac{m R^{2}}{2} + m R^{2}] α$
$\Rightarrow 4 K R^{2} α = \frac{3}{2} m R^{2} α$
$o r α = (\frac{8 K}{3 m}) θ$
$\Rightarrow \vec{α} = - (\frac{8 K}{3 m}) \vec{θ}$
$H e n c e ω^{2} = \frac{8 K}{3 m}$
$o r ω = 2 \sqrt{\frac{2 K}{3 m}}$
$\Rightarrow ω = 2 \sqrt{\frac{2 \times 300}{3 \times 2}} = 20 r a d / s$

PHXI14:OSCILLATIONS

364399 A mass of $0.2 k g$ is attached to the lower end of a massless spring of force constant $200 N / m$ , the upper end of which is fixed to a rigid support Which of the following statement is true?

1 In equilibrium, the spring will be stretched by

1 c m

2 If the mass is raised till the spring becomes unstretched and then released, it will go down by

2 c m

before moving upwards

3 The frequency of oscillation will be nearly

5 H z

4 All the above

Explanation:

$A t e q u i l i b r i u m x = \frac{m g}{k}$
$= \frac{0.2 \times 10}{200} = 0.01 m = 1 c m$
$f = \frac{1}{2 π} \sqrt{\frac{K}{m}} = \frac{1}{2 π} \sqrt{\frac{200}{0.2}} \approx 5.0 H z$

PHXI14:OSCILLATIONS

364400 A mass $m$ is hung on an ideal massless spring. Another equal mass is connected to the other end of the spring. the whole system is at rest. At $t = 0, m$ is released and the system falls freely under gravity. Assume that natural length of the spring is $L_{0}$ , its initial stretched length is $L$ and the acceleration due to gravity is $g$ . What is distance between masses as function of time?
supporting img

1

L_{0} \cos \sqrt{\frac{2 k}{m}} t

2

L_{0} + (L - L_{0}) \cos \sqrt{\frac{2 k}{m}} t

3

L_{0} + (L - L_{0}) \sin \sqrt{\frac{2 k}{m}} t

4

L_{0} \sin \sqrt{\frac{2 k}{m}} t

Explanation:

In $C M$ frame both the masses execute SHM with
$ω = \sqrt{\frac{k}{μ}} = \sqrt{\frac{2 k}{m}}$
Initially particles are at extreme Initial extension in the spring is $(L - L_{o})$
Distance between them $= L_{0} + (L - L_{0}) \cos \sqrt{\frac{2 k}{m}} t$

PHXI14:OSCILLATIONS

364401 A mass $m_{1}$ connected to a horizontal spring perform SHM with amplitude $A$ . While mass $m_{1}$ is passing through mean position, another mass $m_{2}$ is placed on it so that both the masses move together with amplitude $A_{1}$ . The ratio of $\frac{A_{1}}{A}$ is $(m_{2} < m_{1})$

1

{[\frac{m_{1}}{m_{1} + m_{2}}]}^{\frac{1}{2}}

2

{[\frac{m_{1} + m_{2}}{m_{1}}]}^{\frac{1}{2}}

3

{[\frac{m_{2}}{m_{1} + m_{2}}]}^{\frac{1}{2}}

4

{[\frac{m_{1} + m_{2}}{m_{2}}]}^{\frac{1}{2}}

Explanation:

Potential energy of oscillating mass is given by
$P E = \frac{1}{2} k x^{2}$
When only one block is oscillating at $x = A$
$E_{max} = \frac{1}{2} k A^{2}$ and at mean position $x = 0$ . So,
$E = 0$
$∴ A \propto \frac{1}{\sqrt{m}} (1)$
When mass $m_{2}$ is placed on top of the mass $m$ . Then, total mass is $(m_{1} + m_{2})$ and $E = 0$ at this point, as $x = 0$
When they reaches at $x = A$ ,
then $A_{1} \propto \frac{1}{\sqrt{m_{1} + m_{2}}} (2)$
Dividing eq.(2) by (1)
$\frac{A_{1}}{A} = {(\frac{m_{1}}{m_{1} + m_{2}})}^{\frac{1}{2}}$

MHTCET - 2016

PHXI14:OSCILLATIONS

364402 A body of mass $64 g$ is made to oscillate turn by turn on two different springs $A$ and $B$ . Spring $A$ and $B$ has force constant $4 \frac{N}{m}$ and $16 \frac{N}{m}$ respectively. If $T_{1}$ and $T_{2}$ are period of oscillations of spring $A$ and $B$ respectively, then $\frac{T_{1} + T_{2}}{T_{1} - T_{2}}$ will be

1

1 : 2

2

1 : 3

3

3 : 1

4

2 : 1

Explanation:

Given, $m = 64 g = 64 \times 10^{- 3} k g$
$k_{A} = 4 N / m, k_{B} = 16 N / m$
The time period of oscillation of a spring
$T = 2 π \sqrt{\frac{m}{k}}$
$\Rightarrow T_{A} = T_{1} = 2 π \sqrt{\frac{m}{k_{A}}} = 2 π \sqrt{\frac{64 \times 10^{- 3}}{4}} (1)$
$= 2 π \sqrt{16 \times 10^{- 3}}$
$a n d T_{B} = T_{2} = 2 π \sqrt{\frac{m}{k_{B}}} = 2 π \sqrt{\frac{64 \times 10^{- 3}}{16}} (2)$
$= 2 π \sqrt{4 \times 10^{- 3}} (2)$
Dividing eq.(1) by eq.(2) we get
$\begin{aligned} \frac{T_{1}}{T_{2}} = \sqrt{\frac{16 \times 10^{- 3}}{4 \times 10^{- 3}}} = 2 \\ \Rightarrow & T_{1} = 2 T_{2} \\ ∴ \frac{T_{1} + T_{2}}{T_{1} - T_{2}} = \frac{2 T_{2} + T_{2}}{2 T_{2} - T_{2}} = \frac{3 T_{2}}{T_{2}} = \frac{3}{1} or 3 : 1 \end{aligned}$

MHTCET - 2020

PHXI14:OSCILLATIONS

364398 A uniform cylinder of mass $m = 2 k g$ and radius $R = 0.50$ $m$ is in equilibrium on an inclined plane by the action of a light spring of stiffness $K = 300 N / m$ , gravity and reaction force acting on it. If the angle of inclination of the plane is $ϕ = 30^{\circ}$ , find the angular frequency of small oscillation of the cylinder.
supporting img

1

20 r a d / s

2

55 r a d / s

3

29 r a d / s

4

32 r a d / s

Explanation:

Here point of contact $P$ , will be instantaneous centre of rotation. Let the cylinder rotates a small angle $θ$ beyond its equilibrium position about an axis passing through $P$ . Here the increase in length of spring will be $2 R θ$ . Thus the increased spring force provides a restoring torque about $P$ .
supporting img
Now writing torque equation about $P$ as:
$Δ τ_{p} = I_{p} α$
$(K \cdot 2 R θ) \cdot 2 R = [\frac{m R^{2}}{2} + m R^{2}] α$
$\Rightarrow 4 K R^{2} α = \frac{3}{2} m R^{2} α$
$o r α = (\frac{8 K}{3 m}) θ$
$\Rightarrow \vec{α} = - (\frac{8 K}{3 m}) \vec{θ}$
$H e n c e ω^{2} = \frac{8 K}{3 m}$
$o r ω = 2 \sqrt{\frac{2 K}{3 m}}$
$\Rightarrow ω = 2 \sqrt{\frac{2 \times 300}{3 \times 2}} = 20 r a d / s$

PHXI14:OSCILLATIONS

364399 A mass of $0.2 k g$ is attached to the lower end of a massless spring of force constant $200 N / m$ , the upper end of which is fixed to a rigid support Which of the following statement is true?

1 In equilibrium, the spring will be stretched by

1 c m

2 If the mass is raised till the spring becomes unstretched and then released, it will go down by

2 c m

before moving upwards

3 The frequency of oscillation will be nearly

5 H z

4 All the above

Explanation:

$A t e q u i l i b r i u m x = \frac{m g}{k}$
$= \frac{0.2 \times 10}{200} = 0.01 m = 1 c m$
$f = \frac{1}{2 π} \sqrt{\frac{K}{m}} = \frac{1}{2 π} \sqrt{\frac{200}{0.2}} \approx 5.0 H z$

PHXI14:OSCILLATIONS

364400 A mass $m$ is hung on an ideal massless spring. Another equal mass is connected to the other end of the spring. the whole system is at rest. At $t = 0, m$ is released and the system falls freely under gravity. Assume that natural length of the spring is $L_{0}$ , its initial stretched length is $L$ and the acceleration due to gravity is $g$ . What is distance between masses as function of time?
supporting img

1

L_{0} \cos \sqrt{\frac{2 k}{m}} t

2

L_{0} + (L - L_{0}) \cos \sqrt{\frac{2 k}{m}} t

3

L_{0} + (L - L_{0}) \sin \sqrt{\frac{2 k}{m}} t

4

L_{0} \sin \sqrt{\frac{2 k}{m}} t

Explanation:

In $C M$ frame both the masses execute SHM with
$ω = \sqrt{\frac{k}{μ}} = \sqrt{\frac{2 k}{m}}$
Initially particles are at extreme Initial extension in the spring is $(L - L_{o})$
Distance between them $= L_{0} + (L - L_{0}) \cos \sqrt{\frac{2 k}{m}} t$

PHXI14:OSCILLATIONS

364401 A mass $m_{1}$ connected to a horizontal spring perform SHM with amplitude $A$ . While mass $m_{1}$ is passing through mean position, another mass $m_{2}$ is placed on it so that both the masses move together with amplitude $A_{1}$ . The ratio of $\frac{A_{1}}{A}$ is $(m_{2} < m_{1})$

1

{[\frac{m_{1}}{m_{1} + m_{2}}]}^{\frac{1}{2}}

2

{[\frac{m_{1} + m_{2}}{m_{1}}]}^{\frac{1}{2}}

3

{[\frac{m_{2}}{m_{1} + m_{2}}]}^{\frac{1}{2}}

4

{[\frac{m_{1} + m_{2}}{m_{2}}]}^{\frac{1}{2}}

Explanation:

Potential energy of oscillating mass is given by
$P E = \frac{1}{2} k x^{2}$
When only one block is oscillating at $x = A$
$E_{max} = \frac{1}{2} k A^{2}$ and at mean position $x = 0$ . So,
$E = 0$
$∴ A \propto \frac{1}{\sqrt{m}} (1)$
When mass $m_{2}$ is placed on top of the mass $m$ . Then, total mass is $(m_{1} + m_{2})$ and $E = 0$ at this point, as $x = 0$
When they reaches at $x = A$ ,
then $A_{1} \propto \frac{1}{\sqrt{m_{1} + m_{2}}} (2)$
Dividing eq.(2) by (1)
$\frac{A_{1}}{A} = {(\frac{m_{1}}{m_{1} + m_{2}})}^{\frac{1}{2}}$

MHTCET - 2016

PHXI14:OSCILLATIONS

364402 A body of mass $64 g$ is made to oscillate turn by turn on two different springs $A$ and $B$ . Spring $A$ and $B$ has force constant $4 \frac{N}{m}$ and $16 \frac{N}{m}$ respectively. If $T_{1}$ and $T_{2}$ are period of oscillations of spring $A$ and $B$ respectively, then $\frac{T_{1} + T_{2}}{T_{1} - T_{2}}$ will be

1

1 : 2

2

1 : 3

3

3 : 1

4

2 : 1

Explanation:

Given, $m = 64 g = 64 \times 10^{- 3} k g$
$k_{A} = 4 N / m, k_{B} = 16 N / m$
The time period of oscillation of a spring
$T = 2 π \sqrt{\frac{m}{k}}$
$\Rightarrow T_{A} = T_{1} = 2 π \sqrt{\frac{m}{k_{A}}} = 2 π \sqrt{\frac{64 \times 10^{- 3}}{4}} (1)$
$= 2 π \sqrt{16 \times 10^{- 3}}$
$a n d T_{B} = T_{2} = 2 π \sqrt{\frac{m}{k_{B}}} = 2 π \sqrt{\frac{64 \times 10^{- 3}}{16}} (2)$
$= 2 π \sqrt{4 \times 10^{- 3}} (2)$
Dividing eq.(1) by eq.(2) we get
$\begin{aligned} \frac{T_{1}}{T_{2}} = \sqrt{\frac{16 \times 10^{- 3}}{4 \times 10^{- 3}}} = 2 \\ \Rightarrow & T_{1} = 2 T_{2} \\ ∴ \frac{T_{1} + T_{2}}{T_{1} - T_{2}} = \frac{2 T_{2} + T_{2}}{2 T_{2} - T_{2}} = \frac{3 T_{2}}{T_{2}} = \frac{3}{1} or 3 : 1 \end{aligned}$

MHTCET - 2020

PHXI14:OSCILLATIONS

364398 A uniform cylinder of mass $m = 2 k g$ and radius $R = 0.50$ $m$ is in equilibrium on an inclined plane by the action of a light spring of stiffness $K = 300 N / m$ , gravity and reaction force acting on it. If the angle of inclination of the plane is $ϕ = 30^{\circ}$ , find the angular frequency of small oscillation of the cylinder.
supporting img

1

20 r a d / s

2

55 r a d / s

3

29 r a d / s

4

32 r a d / s

Explanation:

Here point of contact $P$ , will be instantaneous centre of rotation. Let the cylinder rotates a small angle $θ$ beyond its equilibrium position about an axis passing through $P$ . Here the increase in length of spring will be $2 R θ$ . Thus the increased spring force provides a restoring torque about $P$ .
supporting img
Now writing torque equation about $P$ as:
$Δ τ_{p} = I_{p} α$
$(K \cdot 2 R θ) \cdot 2 R = [\frac{m R^{2}}{2} + m R^{2}] α$
$\Rightarrow 4 K R^{2} α = \frac{3}{2} m R^{2} α$
$o r α = (\frac{8 K}{3 m}) θ$
$\Rightarrow \vec{α} = - (\frac{8 K}{3 m}) \vec{θ}$
$H e n c e ω^{2} = \frac{8 K}{3 m}$
$o r ω = 2 \sqrt{\frac{2 K}{3 m}}$
$\Rightarrow ω = 2 \sqrt{\frac{2 \times 300}{3 \times 2}} = 20 r a d / s$

PHXI14:OSCILLATIONS

364399 A mass of $0.2 k g$ is attached to the lower end of a massless spring of force constant $200 N / m$ , the upper end of which is fixed to a rigid support Which of the following statement is true?

1 In equilibrium, the spring will be stretched by

1 c m

2 If the mass is raised till the spring becomes unstretched and then released, it will go down by

2 c m

before moving upwards

3 The frequency of oscillation will be nearly

5 H z

4 All the above

Explanation:

$A t e q u i l i b r i u m x = \frac{m g}{k}$
$= \frac{0.2 \times 10}{200} = 0.01 m = 1 c m$
$f = \frac{1}{2 π} \sqrt{\frac{K}{m}} = \frac{1}{2 π} \sqrt{\frac{200}{0.2}} \approx 5.0 H z$

PHXI14:OSCILLATIONS

364400 A mass $m$ is hung on an ideal massless spring. Another equal mass is connected to the other end of the spring. the whole system is at rest. At $t = 0, m$ is released and the system falls freely under gravity. Assume that natural length of the spring is $L_{0}$ , its initial stretched length is $L$ and the acceleration due to gravity is $g$ . What is distance between masses as function of time?
supporting img

1

L_{0} \cos \sqrt{\frac{2 k}{m}} t

2

L_{0} + (L - L_{0}) \cos \sqrt{\frac{2 k}{m}} t

3

L_{0} + (L - L_{0}) \sin \sqrt{\frac{2 k}{m}} t

4

L_{0} \sin \sqrt{\frac{2 k}{m}} t

Explanation:

In $C M$ frame both the masses execute SHM with
$ω = \sqrt{\frac{k}{μ}} = \sqrt{\frac{2 k}{m}}$
Initially particles are at extreme Initial extension in the spring is $(L - L_{o})$
Distance between them $= L_{0} + (L - L_{0}) \cos \sqrt{\frac{2 k}{m}} t$

PHXI14:OSCILLATIONS

364401 A mass $m_{1}$ connected to a horizontal spring perform SHM with amplitude $A$ . While mass $m_{1}$ is passing through mean position, another mass $m_{2}$ is placed on it so that both the masses move together with amplitude $A_{1}$ . The ratio of $\frac{A_{1}}{A}$ is $(m_{2} < m_{1})$

1

{[\frac{m_{1}}{m_{1} + m_{2}}]}^{\frac{1}{2}}

2

{[\frac{m_{1} + m_{2}}{m_{1}}]}^{\frac{1}{2}}

3

{[\frac{m_{2}}{m_{1} + m_{2}}]}^{\frac{1}{2}}

4

{[\frac{m_{1} + m_{2}}{m_{2}}]}^{\frac{1}{2}}

Explanation:

Potential energy of oscillating mass is given by
$P E = \frac{1}{2} k x^{2}$
When only one block is oscillating at $x = A$
$E_{max} = \frac{1}{2} k A^{2}$ and at mean position $x = 0$ . So,
$E = 0$
$∴ A \propto \frac{1}{\sqrt{m}} (1)$
When mass $m_{2}$ is placed on top of the mass $m$ . Then, total mass is $(m_{1} + m_{2})$ and $E = 0$ at this point, as $x = 0$
When they reaches at $x = A$ ,
then $A_{1} \propto \frac{1}{\sqrt{m_{1} + m_{2}}} (2)$
Dividing eq.(2) by (1)
$\frac{A_{1}}{A} = {(\frac{m_{1}}{m_{1} + m_{2}})}^{\frac{1}{2}}$

MHTCET - 2016

PHXI14:OSCILLATIONS

364402 A body of mass $64 g$ is made to oscillate turn by turn on two different springs $A$ and $B$ . Spring $A$ and $B$ has force constant $4 \frac{N}{m}$ and $16 \frac{N}{m}$ respectively. If $T_{1}$ and $T_{2}$ are period of oscillations of spring $A$ and $B$ respectively, then $\frac{T_{1} + T_{2}}{T_{1} - T_{2}}$ will be

1

1 : 2

2

1 : 3

3

3 : 1

4

2 : 1

Explanation:

Given, $m = 64 g = 64 \times 10^{- 3} k g$
$k_{A} = 4 N / m, k_{B} = 16 N / m$
The time period of oscillation of a spring
$T = 2 π \sqrt{\frac{m}{k}}$
$\Rightarrow T_{A} = T_{1} = 2 π \sqrt{\frac{m}{k_{A}}} = 2 π \sqrt{\frac{64 \times 10^{- 3}}{4}} (1)$
$= 2 π \sqrt{16 \times 10^{- 3}}$
$a n d T_{B} = T_{2} = 2 π \sqrt{\frac{m}{k_{B}}} = 2 π \sqrt{\frac{64 \times 10^{- 3}}{16}} (2)$
$= 2 π \sqrt{4 \times 10^{- 3}} (2)$
Dividing eq.(1) by eq.(2) we get
$\begin{aligned} \frac{T_{1}}{T_{2}} = \sqrt{\frac{16 \times 10^{- 3}}{4 \times 10^{- 3}}} = 2 \\ \Rightarrow & T_{1} = 2 T_{2} \\ ∴ \frac{T_{1} + T_{2}}{T_{1} - T_{2}} = \frac{2 T_{2} + T_{2}}{2 T_{2} - T_{2}} = \frac{3 T_{2}}{T_{2}} = \frac{3}{1} or 3 : 1 \end{aligned}$

MHTCET - 2020

PHXI14:OSCILLATIONS

364398 A uniform cylinder of mass $m = 2 k g$ and radius $R = 0.50$ $m$ is in equilibrium on an inclined plane by the action of a light spring of stiffness $K = 300 N / m$ , gravity and reaction force acting on it. If the angle of inclination of the plane is $ϕ = 30^{\circ}$ , find the angular frequency of small oscillation of the cylinder.
supporting img

1

20 r a d / s

2

55 r a d / s

3

29 r a d / s

4

32 r a d / s

Explanation:

Here point of contact $P$ , will be instantaneous centre of rotation. Let the cylinder rotates a small angle $θ$ beyond its equilibrium position about an axis passing through $P$ . Here the increase in length of spring will be $2 R θ$ . Thus the increased spring force provides a restoring torque about $P$ .
supporting img
Now writing torque equation about $P$ as:
$Δ τ_{p} = I_{p} α$
$(K \cdot 2 R θ) \cdot 2 R = [\frac{m R^{2}}{2} + m R^{2}] α$
$\Rightarrow 4 K R^{2} α = \frac{3}{2} m R^{2} α$
$o r α = (\frac{8 K}{3 m}) θ$
$\Rightarrow \vec{α} = - (\frac{8 K}{3 m}) \vec{θ}$
$H e n c e ω^{2} = \frac{8 K}{3 m}$
$o r ω = 2 \sqrt{\frac{2 K}{3 m}}$
$\Rightarrow ω = 2 \sqrt{\frac{2 \times 300}{3 \times 2}} = 20 r a d / s$

PHXI14:OSCILLATIONS

364399 A mass of $0.2 k g$ is attached to the lower end of a massless spring of force constant $200 N / m$ , the upper end of which is fixed to a rigid support Which of the following statement is true?

1 In equilibrium, the spring will be stretched by

1 c m

2 If the mass is raised till the spring becomes unstretched and then released, it will go down by

2 c m

before moving upwards

3 The frequency of oscillation will be nearly

5 H z

4 All the above

Explanation:

$A t e q u i l i b r i u m x = \frac{m g}{k}$
$= \frac{0.2 \times 10}{200} = 0.01 m = 1 c m$
$f = \frac{1}{2 π} \sqrt{\frac{K}{m}} = \frac{1}{2 π} \sqrt{\frac{200}{0.2}} \approx 5.0 H z$

PHXI14:OSCILLATIONS

364400 A mass $m$ is hung on an ideal massless spring. Another equal mass is connected to the other end of the spring. the whole system is at rest. At $t = 0, m$ is released and the system falls freely under gravity. Assume that natural length of the spring is $L_{0}$ , its initial stretched length is $L$ and the acceleration due to gravity is $g$ . What is distance between masses as function of time?
supporting img

1

L_{0} \cos \sqrt{\frac{2 k}{m}} t

2

L_{0} + (L - L_{0}) \cos \sqrt{\frac{2 k}{m}} t

3

L_{0} + (L - L_{0}) \sin \sqrt{\frac{2 k}{m}} t

4

L_{0} \sin \sqrt{\frac{2 k}{m}} t

Explanation:

In $C M$ frame both the masses execute SHM with
$ω = \sqrt{\frac{k}{μ}} = \sqrt{\frac{2 k}{m}}$
Initially particles are at extreme Initial extension in the spring is $(L - L_{o})$
Distance between them $= L_{0} + (L - L_{0}) \cos \sqrt{\frac{2 k}{m}} t$

PHXI14:OSCILLATIONS

364401 A mass $m_{1}$ connected to a horizontal spring perform SHM with amplitude $A$ . While mass $m_{1}$ is passing through mean position, another mass $m_{2}$ is placed on it so that both the masses move together with amplitude $A_{1}$ . The ratio of $\frac{A_{1}}{A}$ is $(m_{2} < m_{1})$

1

{[\frac{m_{1}}{m_{1} + m_{2}}]}^{\frac{1}{2}}

2

{[\frac{m_{1} + m_{2}}{m_{1}}]}^{\frac{1}{2}}

3

{[\frac{m_{2}}{m_{1} + m_{2}}]}^{\frac{1}{2}}

4

{[\frac{m_{1} + m_{2}}{m_{2}}]}^{\frac{1}{2}}

Explanation:

Potential energy of oscillating mass is given by
$P E = \frac{1}{2} k x^{2}$
When only one block is oscillating at $x = A$
$E_{max} = \frac{1}{2} k A^{2}$ and at mean position $x = 0$ . So,
$E = 0$
$∴ A \propto \frac{1}{\sqrt{m}} (1)$
When mass $m_{2}$ is placed on top of the mass $m$ . Then, total mass is $(m_{1} + m_{2})$ and $E = 0$ at this point, as $x = 0$
When they reaches at $x = A$ ,
then $A_{1} \propto \frac{1}{\sqrt{m_{1} + m_{2}}} (2)$
Dividing eq.(2) by (1)
$\frac{A_{1}}{A} = {(\frac{m_{1}}{m_{1} + m_{2}})}^{\frac{1}{2}}$

MHTCET - 2016

PHXI14:OSCILLATIONS

364402 A body of mass $64 g$ is made to oscillate turn by turn on two different springs $A$ and $B$ . Spring $A$ and $B$ has force constant $4 \frac{N}{m}$ and $16 \frac{N}{m}$ respectively. If $T_{1}$ and $T_{2}$ are period of oscillations of spring $A$ and $B$ respectively, then $\frac{T_{1} + T_{2}}{T_{1} - T_{2}}$ will be

1

1 : 2

2

1 : 3

3

3 : 1

4

2 : 1

Explanation:

Given, $m = 64 g = 64 \times 10^{- 3} k g$
$k_{A} = 4 N / m, k_{B} = 16 N / m$
The time period of oscillation of a spring
$T = 2 π \sqrt{\frac{m}{k}}$
$\Rightarrow T_{A} = T_{1} = 2 π \sqrt{\frac{m}{k_{A}}} = 2 π \sqrt{\frac{64 \times 10^{- 3}}{4}} (1)$
$= 2 π \sqrt{16 \times 10^{- 3}}$
$a n d T_{B} = T_{2} = 2 π \sqrt{\frac{m}{k_{B}}} = 2 π \sqrt{\frac{64 \times 10^{- 3}}{16}} (2)$
$= 2 π \sqrt{4 \times 10^{- 3}} (2)$
Dividing eq.(1) by eq.(2) we get
$\begin{aligned} \frac{T_{1}}{T_{2}} = \sqrt{\frac{16 \times 10^{- 3}}{4 \times 10^{- 3}}} = 2 \\ \Rightarrow & T_{1} = 2 T_{2} \\ ∴ \frac{T_{1} + T_{2}}{T_{1} - T_{2}} = \frac{2 T_{2} + T_{2}}{2 T_{2} - T_{2}} = \frac{3 T_{2}}{T_{2}} = \frac{3}{1} or 3 : 1 \end{aligned}$

MHTCET - 2020

PHXI14:OSCILLATIONS

364398 A uniform cylinder of mass $m = 2 k g$ and radius $R = 0.50$ $m$ is in equilibrium on an inclined plane by the action of a light spring of stiffness $K = 300 N / m$ , gravity and reaction force acting on it. If the angle of inclination of the plane is $ϕ = 30^{\circ}$ , find the angular frequency of small oscillation of the cylinder.
supporting img

1

20 r a d / s

2

55 r a d / s

3

29 r a d / s

4

32 r a d / s

Explanation:

Here point of contact $P$ , will be instantaneous centre of rotation. Let the cylinder rotates a small angle $θ$ beyond its equilibrium position about an axis passing through $P$ . Here the increase in length of spring will be $2 R θ$ . Thus the increased spring force provides a restoring torque about $P$ .
supporting img
Now writing torque equation about $P$ as:
$Δ τ_{p} = I_{p} α$
$(K \cdot 2 R θ) \cdot 2 R = [\frac{m R^{2}}{2} + m R^{2}] α$
$\Rightarrow 4 K R^{2} α = \frac{3}{2} m R^{2} α$
$o r α = (\frac{8 K}{3 m}) θ$
$\Rightarrow \vec{α} = - (\frac{8 K}{3 m}) \vec{θ}$
$H e n c e ω^{2} = \frac{8 K}{3 m}$
$o r ω = 2 \sqrt{\frac{2 K}{3 m}}$
$\Rightarrow ω = 2 \sqrt{\frac{2 \times 300}{3 \times 2}} = 20 r a d / s$

PHXI14:OSCILLATIONS

364399 A mass of $0.2 k g$ is attached to the lower end of a massless spring of force constant $200 N / m$ , the upper end of which is fixed to a rigid support Which of the following statement is true?

1 In equilibrium, the spring will be stretched by

1 c m

2 If the mass is raised till the spring becomes unstretched and then released, it will go down by

2 c m

before moving upwards

3 The frequency of oscillation will be nearly

5 H z

4 All the above

Explanation:

$A t e q u i l i b r i u m x = \frac{m g}{k}$
$= \frac{0.2 \times 10}{200} = 0.01 m = 1 c m$
$f = \frac{1}{2 π} \sqrt{\frac{K}{m}} = \frac{1}{2 π} \sqrt{\frac{200}{0.2}} \approx 5.0 H z$

PHXI14:OSCILLATIONS

364400 A mass $m$ is hung on an ideal massless spring. Another equal mass is connected to the other end of the spring. the whole system is at rest. At $t = 0, m$ is released and the system falls freely under gravity. Assume that natural length of the spring is $L_{0}$ , its initial stretched length is $L$ and the acceleration due to gravity is $g$ . What is distance between masses as function of time?
supporting img

1

L_{0} \cos \sqrt{\frac{2 k}{m}} t

2

L_{0} + (L - L_{0}) \cos \sqrt{\frac{2 k}{m}} t

3

L_{0} + (L - L_{0}) \sin \sqrt{\frac{2 k}{m}} t

4

L_{0} \sin \sqrt{\frac{2 k}{m}} t

Explanation:

In $C M$ frame both the masses execute SHM with
$ω = \sqrt{\frac{k}{μ}} = \sqrt{\frac{2 k}{m}}$
Initially particles are at extreme Initial extension in the spring is $(L - L_{o})$
Distance between them $= L_{0} + (L - L_{0}) \cos \sqrt{\frac{2 k}{m}} t$

PHXI14:OSCILLATIONS

364401 A mass $m_{1}$ connected to a horizontal spring perform SHM with amplitude $A$ . While mass $m_{1}$ is passing through mean position, another mass $m_{2}$ is placed on it so that both the masses move together with amplitude $A_{1}$ . The ratio of $\frac{A_{1}}{A}$ is $(m_{2} < m_{1})$

1

{[\frac{m_{1}}{m_{1} + m_{2}}]}^{\frac{1}{2}}

2

{[\frac{m_{1} + m_{2}}{m_{1}}]}^{\frac{1}{2}}

3

{[\frac{m_{2}}{m_{1} + m_{2}}]}^{\frac{1}{2}}

4

{[\frac{m_{1} + m_{2}}{m_{2}}]}^{\frac{1}{2}}

Explanation:

Potential energy of oscillating mass is given by
$P E = \frac{1}{2} k x^{2}$
When only one block is oscillating at $x = A$
$E_{max} = \frac{1}{2} k A^{2}$ and at mean position $x = 0$ . So,
$E = 0$
$∴ A \propto \frac{1}{\sqrt{m}} (1)$
When mass $m_{2}$ is placed on top of the mass $m$ . Then, total mass is $(m_{1} + m_{2})$ and $E = 0$ at this point, as $x = 0$
When they reaches at $x = A$ ,
then $A_{1} \propto \frac{1}{\sqrt{m_{1} + m_{2}}} (2)$
Dividing eq.(2) by (1)
$\frac{A_{1}}{A} = {(\frac{m_{1}}{m_{1} + m_{2}})}^{\frac{1}{2}}$

MHTCET - 2016

PHXI14:OSCILLATIONS

364402 A body of mass $64 g$ is made to oscillate turn by turn on two different springs $A$ and $B$ . Spring $A$ and $B$ has force constant $4 \frac{N}{m}$ and $16 \frac{N}{m}$ respectively. If $T_{1}$ and $T_{2}$ are period of oscillations of spring $A$ and $B$ respectively, then $\frac{T_{1} + T_{2}}{T_{1} - T_{2}}$ will be

1

1 : 2

2

1 : 3

3

3 : 1

4

2 : 1

Explanation:

Given, $m = 64 g = 64 \times 10^{- 3} k g$
$k_{A} = 4 N / m, k_{B} = 16 N / m$
The time period of oscillation of a spring
$T = 2 π \sqrt{\frac{m}{k}}$
$\Rightarrow T_{A} = T_{1} = 2 π \sqrt{\frac{m}{k_{A}}} = 2 π \sqrt{\frac{64 \times 10^{- 3}}{4}} (1)$
$= 2 π \sqrt{16 \times 10^{- 3}}$
$a n d T_{B} = T_{2} = 2 π \sqrt{\frac{m}{k_{B}}} = 2 π \sqrt{\frac{64 \times 10^{- 3}}{16}} (2)$
$= 2 π \sqrt{4 \times 10^{- 3}} (2)$
Dividing eq.(1) by eq.(2) we get
$\begin{aligned} \frac{T_{1}}{T_{2}} = \sqrt{\frac{16 \times 10^{- 3}}{4 \times 10^{- 3}}} = 2 \\ \Rightarrow & T_{1} = 2 T_{2} \\ ∴ \frac{T_{1} + T_{2}}{T_{1} - T_{2}} = \frac{2 T_{2} + T_{2}}{2 T_{2} - T_{2}} = \frac{3 T_{2}}{T_{2}} = \frac{3}{1} or 3 : 1 \end{aligned}$

MHTCET - 2020

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