QBManager

PHXI14:OSCILLATIONS

364407 A mass of $250 g$ hangs on a spring and oscillates vertically with a period of $1.1 s$ . To double the period, what mass must be added to the $250 g$ ? (Ignore the mass of the spring)

1

250 g

2

450 g

3

750 g

4

550 g

Explanation:

As, $T \propto \sqrt{m} \Rightarrow 1.1 \propto \sqrt{250}$
and $\Rightarrow 2.2 \propto \sqrt{250 + x}$
$∴ \frac{1}{2} = \sqrt{\frac{250}{250 + x}} \Rightarrow x = 750 g$

PHXI14:OSCILLATIONS

364408 A block of mass $m$ rests on a platform. The platform is given up and down SHM with an amplitude $d$ . What can be the maximum frequency so that the block never leaves the platform?

1

\frac{1}{2 π} \sqrt{g / d}

2

\sqrt{g / d}

3

2 π \sqrt{g / d}

4

\frac{1}{2 π} (g / d)

Explanation:

For the gievn amplitude $d$ the block looses contact with the platform when maximum acceleration is equal to $g$ at the highest position.
$\begin{aligned} k d = m g \Rightarrow m ω^{2} d = m g \\ \Rightarrow ω = \sqrt{\frac{g}{d}} \Rightarrow n = \frac{1}{2 π} \sqrt{\frac{g}{d}} \end{aligned}$

PHXI14:OSCILLATIONS

364409 For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1 k g$ , the angular
supporting img
frequency is $ω_{1}$ . When the mass is $2 k g$ the angular frequency is $ω_{2}$ . The ratio $ω_{2} / ω_{1}$ is

1 2

2

\sqrt{2}

3

\frac{1}{2}

4

\frac{1}{\sqrt{2}}

Explanation:

As, for osicillations of a spring - mass system,
$T = 2 π \sqrt{\frac{m}{k}}$ and $T = \frac{2 π}{ω}$
$\frac{2 π}{ω} = 2 π \sqrt{\frac{m}{k}}; ω^{2} = \frac{k}{m}$
or $ω_{1}^{2} = \frac{k}{m} = \frac{k}{1} (1)$
and $ω_{2}^{2} = \frac{k}{m_{2}} = \frac{k}{2} (2)$
Dividing equations (1) and (2)
$∴ \frac{ω_{2}^{2}}{ω_{1}^{2}} = \frac{k / 2}{k / 1} = \frac{1}{2}; \frac{ω_{2}}{ω_{1}} = \frac{1}{\sqrt{2}}$
$[A s, m_{1} = 1 k g, m_{2} = 2 k g]$

JEE - 2023

PHXI14:OSCILLATIONS

364410 A block of mass $100 g$ attached to a spring of spring constant $100 N / m$ is lying on a frictionless floor as shown. The block is moved to compress the spring by $10 c m$ and then released. If the collisions with the wall in front are elastic, then the time period of the motion is: (Given $π^{2} = 10$ )
supporting img

1

0.2 s

2

0.1 s

3

0.15 s

4

0.132 s

Explanation:

Total phase traversed
$= \frac{π}{2} + \frac{π}{2} + \frac{π}{6} + \frac{π}{6} = \frac{4 π}{3}$
time $= \frac{2}{3} T = 0.133 \sec$

PHXI14:OSCILLATIONS

364407 A mass of $250 g$ hangs on a spring and oscillates vertically with a period of $1.1 s$ . To double the period, what mass must be added to the $250 g$ ? (Ignore the mass of the spring)

1

250 g

2

450 g

3

750 g

4

550 g

Explanation:

As, $T \propto \sqrt{m} \Rightarrow 1.1 \propto \sqrt{250}$
and $\Rightarrow 2.2 \propto \sqrt{250 + x}$
$∴ \frac{1}{2} = \sqrt{\frac{250}{250 + x}} \Rightarrow x = 750 g$

PHXI14:OSCILLATIONS

364408 A block of mass $m$ rests on a platform. The platform is given up and down SHM with an amplitude $d$ . What can be the maximum frequency so that the block never leaves the platform?

1

\frac{1}{2 π} \sqrt{g / d}

2

\sqrt{g / d}

3

2 π \sqrt{g / d}

4

\frac{1}{2 π} (g / d)

Explanation:

For the gievn amplitude $d$ the block looses contact with the platform when maximum acceleration is equal to $g$ at the highest position.
$\begin{aligned} k d = m g \Rightarrow m ω^{2} d = m g \\ \Rightarrow ω = \sqrt{\frac{g}{d}} \Rightarrow n = \frac{1}{2 π} \sqrt{\frac{g}{d}} \end{aligned}$

PHXI14:OSCILLATIONS

364409 For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1 k g$ , the angular
supporting img
frequency is $ω_{1}$ . When the mass is $2 k g$ the angular frequency is $ω_{2}$ . The ratio $ω_{2} / ω_{1}$ is

1 2

2

\sqrt{2}

3

\frac{1}{2}

4

\frac{1}{\sqrt{2}}

Explanation:

As, for osicillations of a spring - mass system,
$T = 2 π \sqrt{\frac{m}{k}}$ and $T = \frac{2 π}{ω}$
$\frac{2 π}{ω} = 2 π \sqrt{\frac{m}{k}}; ω^{2} = \frac{k}{m}$
or $ω_{1}^{2} = \frac{k}{m} = \frac{k}{1} (1)$
and $ω_{2}^{2} = \frac{k}{m_{2}} = \frac{k}{2} (2)$
Dividing equations (1) and (2)
$∴ \frac{ω_{2}^{2}}{ω_{1}^{2}} = \frac{k / 2}{k / 1} = \frac{1}{2}; \frac{ω_{2}}{ω_{1}} = \frac{1}{\sqrt{2}}$
$[A s, m_{1} = 1 k g, m_{2} = 2 k g]$

JEE - 2023

PHXI14:OSCILLATIONS

364410 A block of mass $100 g$ attached to a spring of spring constant $100 N / m$ is lying on a frictionless floor as shown. The block is moved to compress the spring by $10 c m$ and then released. If the collisions with the wall in front are elastic, then the time period of the motion is: (Given $π^{2} = 10$ )
supporting img

1

0.2 s

2

0.1 s

3

0.15 s

4

0.132 s

Explanation:

Total phase traversed
$= \frac{π}{2} + \frac{π}{2} + \frac{π}{6} + \frac{π}{6} = \frac{4 π}{3}$
time $= \frac{2}{3} T = 0.133 \sec$

PHXI14:OSCILLATIONS

364407 A mass of $250 g$ hangs on a spring and oscillates vertically with a period of $1.1 s$ . To double the period, what mass must be added to the $250 g$ ? (Ignore the mass of the spring)

1

250 g

2

450 g

3

750 g

4

550 g

Explanation:

As, $T \propto \sqrt{m} \Rightarrow 1.1 \propto \sqrt{250}$
and $\Rightarrow 2.2 \propto \sqrt{250 + x}$
$∴ \frac{1}{2} = \sqrt{\frac{250}{250 + x}} \Rightarrow x = 750 g$

PHXI14:OSCILLATIONS

364408 A block of mass $m$ rests on a platform. The platform is given up and down SHM with an amplitude $d$ . What can be the maximum frequency so that the block never leaves the platform?

1

\frac{1}{2 π} \sqrt{g / d}

2

\sqrt{g / d}

3

2 π \sqrt{g / d}

4

\frac{1}{2 π} (g / d)

Explanation:

For the gievn amplitude $d$ the block looses contact with the platform when maximum acceleration is equal to $g$ at the highest position.
$\begin{aligned} k d = m g \Rightarrow m ω^{2} d = m g \\ \Rightarrow ω = \sqrt{\frac{g}{d}} \Rightarrow n = \frac{1}{2 π} \sqrt{\frac{g}{d}} \end{aligned}$

PHXI14:OSCILLATIONS

364409 For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1 k g$ , the angular
supporting img
frequency is $ω_{1}$ . When the mass is $2 k g$ the angular frequency is $ω_{2}$ . The ratio $ω_{2} / ω_{1}$ is

1 2

2

\sqrt{2}

3

\frac{1}{2}

4

\frac{1}{\sqrt{2}}

Explanation:

As, for osicillations of a spring - mass system,
$T = 2 π \sqrt{\frac{m}{k}}$ and $T = \frac{2 π}{ω}$
$\frac{2 π}{ω} = 2 π \sqrt{\frac{m}{k}}; ω^{2} = \frac{k}{m}$
or $ω_{1}^{2} = \frac{k}{m} = \frac{k}{1} (1)$
and $ω_{2}^{2} = \frac{k}{m_{2}} = \frac{k}{2} (2)$
Dividing equations (1) and (2)
$∴ \frac{ω_{2}^{2}}{ω_{1}^{2}} = \frac{k / 2}{k / 1} = \frac{1}{2}; \frac{ω_{2}}{ω_{1}} = \frac{1}{\sqrt{2}}$
$[A s, m_{1} = 1 k g, m_{2} = 2 k g]$

JEE - 2023

PHXI14:OSCILLATIONS

364410 A block of mass $100 g$ attached to a spring of spring constant $100 N / m$ is lying on a frictionless floor as shown. The block is moved to compress the spring by $10 c m$ and then released. If the collisions with the wall in front are elastic, then the time period of the motion is: (Given $π^{2} = 10$ )
supporting img

1

0.2 s

2

0.1 s

3

0.15 s

4

0.132 s

Explanation:

Total phase traversed
$= \frac{π}{2} + \frac{π}{2} + \frac{π}{6} + \frac{π}{6} = \frac{4 π}{3}$
time $= \frac{2}{3} T = 0.133 \sec$

PHXI14:OSCILLATIONS

364407 A mass of $250 g$ hangs on a spring and oscillates vertically with a period of $1.1 s$ . To double the period, what mass must be added to the $250 g$ ? (Ignore the mass of the spring)

1

250 g

2

450 g

3

750 g

4

550 g

Explanation:

As, $T \propto \sqrt{m} \Rightarrow 1.1 \propto \sqrt{250}$
and $\Rightarrow 2.2 \propto \sqrt{250 + x}$
$∴ \frac{1}{2} = \sqrt{\frac{250}{250 + x}} \Rightarrow x = 750 g$

PHXI14:OSCILLATIONS

364408 A block of mass $m$ rests on a platform. The platform is given up and down SHM with an amplitude $d$ . What can be the maximum frequency so that the block never leaves the platform?

1

\frac{1}{2 π} \sqrt{g / d}

2

\sqrt{g / d}

3

2 π \sqrt{g / d}

4

\frac{1}{2 π} (g / d)

Explanation:

For the gievn amplitude $d$ the block looses contact with the platform when maximum acceleration is equal to $g$ at the highest position.
$\begin{aligned} k d = m g \Rightarrow m ω^{2} d = m g \\ \Rightarrow ω = \sqrt{\frac{g}{d}} \Rightarrow n = \frac{1}{2 π} \sqrt{\frac{g}{d}} \end{aligned}$

PHXI14:OSCILLATIONS

364409 For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1 k g$ , the angular
supporting img
frequency is $ω_{1}$ . When the mass is $2 k g$ the angular frequency is $ω_{2}$ . The ratio $ω_{2} / ω_{1}$ is

1 2

2

\sqrt{2}

3

\frac{1}{2}

4

\frac{1}{\sqrt{2}}

Explanation:

As, for osicillations of a spring - mass system,
$T = 2 π \sqrt{\frac{m}{k}}$ and $T = \frac{2 π}{ω}$
$\frac{2 π}{ω} = 2 π \sqrt{\frac{m}{k}}; ω^{2} = \frac{k}{m}$
or $ω_{1}^{2} = \frac{k}{m} = \frac{k}{1} (1)$
and $ω_{2}^{2} = \frac{k}{m_{2}} = \frac{k}{2} (2)$
Dividing equations (1) and (2)
$∴ \frac{ω_{2}^{2}}{ω_{1}^{2}} = \frac{k / 2}{k / 1} = \frac{1}{2}; \frac{ω_{2}}{ω_{1}} = \frac{1}{\sqrt{2}}$
$[A s, m_{1} = 1 k g, m_{2} = 2 k g]$

JEE - 2023

PHXI14:OSCILLATIONS

364410 A block of mass $100 g$ attached to a spring of spring constant $100 N / m$ is lying on a frictionless floor as shown. The block is moved to compress the spring by $10 c m$ and then released. If the collisions with the wall in front are elastic, then the time period of the motion is: (Given $π^{2} = 10$ )
supporting img

1

0.2 s

2

0.1 s

3

0.15 s

4

0.132 s

Explanation:

Total phase traversed
$= \frac{π}{2} + \frac{π}{2} + \frac{π}{6} + \frac{π}{6} = \frac{4 π}{3}$
time $= \frac{2}{3} T = 0.133 \sec$