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PHXI14:OSCILLATIONS

364385 An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $15 c m / \sec$ and the time $\frac{π}{5} \sec$ , find its amplitude

1 2.0

2 3.0

3 1.0

4 1.5

Explanation:

Maximum velocity, $v_{max} = a ω$
$o r v_{max} = a (\frac{2 π}{T})$
$a = \frac{v_{max} T}{2 π} = \frac{15 \times 3.14}{2 \times 3.14 \times 5} = 1.5 c m$

PHXI14:OSCILLATIONS

364386 Two identical particles each of mass $m$ are interconnected by a light spring of stiffness $k$ , the time period for small oscillation is equal to:
supporting img

1

\frac{π}{2} \sqrt{\frac{m}{k}}

2

2 π \sqrt{\frac{m}{k}}

3

π \sqrt{\frac{2 m}{k}}

4

π \sqrt{\frac{m}{2 k}}

Explanation:

The reduced mass
$μ = \frac{(m) (m)}{m + m} = \frac{m}{2}$
The given system is equivalent to a system of a particle of mass $m / 2$ connected to a spring of stiffness $k$ ragidly.
The required period of oscillation.
$T = 2 π \sqrt{\frac{μ}{k}} = 2 π \sqrt{\frac{m / 2}{k}} = π \sqrt{\frac{2 m}{k}}$

PHXI14:OSCILLATIONS

364387 A mass $M$ is suspended with a light spring. An additional mass $m$ added displaces the spring further by a distance $x$ Now, the combined mass will oscillate on the spring with period

1

T = 2 π \sqrt{\frac{m g}{x (M + m)}}

2

T = 2 π \sqrt{\frac{(M + m) x}{m g}}

3

T = \frac{π}{2} \sqrt{\frac{m g}{x (M + m)}}

4

T = 2 π \sqrt{\frac{M + m}{m g x}}

Explanation:

$M g = k x_{1} (1)$
$(M + m) g = k x_{2} (2)$
On substracting Eq. (1) from Eq. (2),
$\begin{aligned} m g = k (x_{2} - x_{1}) \\ k = \frac{m g}{x} (∵ x_{2} - x_{1} = x) \end{aligned}$
Time period $T = 2 π \sqrt{\frac{(M + m)}{k}}$
$= 2 π \sqrt{\frac{(M + m) x}{m g}}$

PHXI14:OSCILLATIONS

364388 A block of mass $M$ is suspended from a light spring of force constant $k$ . Another mass $m$ moving upwards with velocity $v$ shifts the mass $M$ and gets embeded in it. What will be the amplitude of the combined mass?

1

\frac{m v}{\sqrt{(M - m) k}}

2

\frac{m v}{k \sqrt{(M - m)}}

3

\frac{m v}{\sqrt{(M + m) k}}

4

\frac{m v}{k \sqrt{(M + m)}}

Explanation:

If $v^{'}$ are the velocities of the block of mass $M$ and $(M + m)$ while passing from the mean position when executing SHM.
Using law of conservation of linear momentum, we have
$\begin{aligned} m v & = (M + m) v^{'} \\ or v^{'} & = m v / (M + m) \end{aligned}$
Also, maximum $P E =$ maximum $K E$
$\begin{matrix} ∴ \frac{1}{2} k A^{' 2} = \frac{1}{2} (M + m) v^{' 2} \\ A^{'} = {(\frac{M + m}{k})}^{1 / 2} \times \frac{m v}{(M + m)} = \frac{m v}{\sqrt{(M + m) k}} \end{matrix}$

PHXI14:OSCILLATIONS

364389 The masses $m_{1}$ and $m_{2}$ are suspended together by a massless spring of constant $k$ . When the masses are in equilibrium, $m_{1}$ is removed without disturbing the system; the amplitude of vibration is -

1

m_{2} g / k

2

m_{1} g / k

3

\frac{(m_{2} - m_{1}) g}{k}

4

\frac{(m_{1} + m_{2}) g}{k}

Explanation:

In equilibrium $F = (m_{1} + m_{2}) g$
By removing mass $m_{1}$
$\begin{aligned} F^{'} = m_{2} g \\ F_{r} = F^{'} - F = m_{2} g - (m_{1} + m_{2}) g \\ k A = m_{1} g \Rightarrow A = \frac{m_{1} g}{k} \end{aligned}$
since the lower spring exerts force on the block only in the downward motion.

PHXI14:OSCILLATIONS

364385 An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $15 c m / \sec$ and the time $\frac{π}{5} \sec$ , find its amplitude

1 2.0

2 3.0

3 1.0

4 1.5

Explanation:

Maximum velocity, $v_{max} = a ω$
$o r v_{max} = a (\frac{2 π}{T})$
$a = \frac{v_{max} T}{2 π} = \frac{15 \times 3.14}{2 \times 3.14 \times 5} = 1.5 c m$

PHXI14:OSCILLATIONS

364386 Two identical particles each of mass $m$ are interconnected by a light spring of stiffness $k$ , the time period for small oscillation is equal to:
supporting img

1

\frac{π}{2} \sqrt{\frac{m}{k}}

2

2 π \sqrt{\frac{m}{k}}

3

π \sqrt{\frac{2 m}{k}}

4

π \sqrt{\frac{m}{2 k}}

Explanation:

The reduced mass
$μ = \frac{(m) (m)}{m + m} = \frac{m}{2}$
The given system is equivalent to a system of a particle of mass $m / 2$ connected to a spring of stiffness $k$ ragidly.
The required period of oscillation.
$T = 2 π \sqrt{\frac{μ}{k}} = 2 π \sqrt{\frac{m / 2}{k}} = π \sqrt{\frac{2 m}{k}}$

PHXI14:OSCILLATIONS

364387 A mass $M$ is suspended with a light spring. An additional mass $m$ added displaces the spring further by a distance $x$ Now, the combined mass will oscillate on the spring with period

1

T = 2 π \sqrt{\frac{m g}{x (M + m)}}

2

T = 2 π \sqrt{\frac{(M + m) x}{m g}}

3

T = \frac{π}{2} \sqrt{\frac{m g}{x (M + m)}}

4

T = 2 π \sqrt{\frac{M + m}{m g x}}

Explanation:

$M g = k x_{1} (1)$
$(M + m) g = k x_{2} (2)$
On substracting Eq. (1) from Eq. (2),
$\begin{aligned} m g = k (x_{2} - x_{1}) \\ k = \frac{m g}{x} (∵ x_{2} - x_{1} = x) \end{aligned}$
Time period $T = 2 π \sqrt{\frac{(M + m)}{k}}$
$= 2 π \sqrt{\frac{(M + m) x}{m g}}$

PHXI14:OSCILLATIONS

364388 A block of mass $M$ is suspended from a light spring of force constant $k$ . Another mass $m$ moving upwards with velocity $v$ shifts the mass $M$ and gets embeded in it. What will be the amplitude of the combined mass?

1

\frac{m v}{\sqrt{(M - m) k}}

2

\frac{m v}{k \sqrt{(M - m)}}

3

\frac{m v}{\sqrt{(M + m) k}}

4

\frac{m v}{k \sqrt{(M + m)}}

Explanation:

If $v^{'}$ are the velocities of the block of mass $M$ and $(M + m)$ while passing from the mean position when executing SHM.
Using law of conservation of linear momentum, we have
$\begin{aligned} m v & = (M + m) v^{'} \\ or v^{'} & = m v / (M + m) \end{aligned}$
Also, maximum $P E =$ maximum $K E$
$\begin{matrix} ∴ \frac{1}{2} k A^{' 2} = \frac{1}{2} (M + m) v^{' 2} \\ A^{'} = {(\frac{M + m}{k})}^{1 / 2} \times \frac{m v}{(M + m)} = \frac{m v}{\sqrt{(M + m) k}} \end{matrix}$

PHXI14:OSCILLATIONS

364389 The masses $m_{1}$ and $m_{2}$ are suspended together by a massless spring of constant $k$ . When the masses are in equilibrium, $m_{1}$ is removed without disturbing the system; the amplitude of vibration is -

1

m_{2} g / k

2

m_{1} g / k

3

\frac{(m_{2} - m_{1}) g}{k}

4

\frac{(m_{1} + m_{2}) g}{k}

Explanation:

In equilibrium $F = (m_{1} + m_{2}) g$
By removing mass $m_{1}$
$\begin{aligned} F^{'} = m_{2} g \\ F_{r} = F^{'} - F = m_{2} g - (m_{1} + m_{2}) g \\ k A = m_{1} g \Rightarrow A = \frac{m_{1} g}{k} \end{aligned}$
since the lower spring exerts force on the block only in the downward motion.

PHXI14:OSCILLATIONS

364385 An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $15 c m / \sec$ and the time $\frac{π}{5} \sec$ , find its amplitude

1 2.0

2 3.0

3 1.0

4 1.5

Explanation:

Maximum velocity, $v_{max} = a ω$
$o r v_{max} = a (\frac{2 π}{T})$
$a = \frac{v_{max} T}{2 π} = \frac{15 \times 3.14}{2 \times 3.14 \times 5} = 1.5 c m$

PHXI14:OSCILLATIONS

364386 Two identical particles each of mass $m$ are interconnected by a light spring of stiffness $k$ , the time period for small oscillation is equal to:
supporting img

1

\frac{π}{2} \sqrt{\frac{m}{k}}

2

2 π \sqrt{\frac{m}{k}}

3

π \sqrt{\frac{2 m}{k}}

4

π \sqrt{\frac{m}{2 k}}

Explanation:

The reduced mass
$μ = \frac{(m) (m)}{m + m} = \frac{m}{2}$
The given system is equivalent to a system of a particle of mass $m / 2$ connected to a spring of stiffness $k$ ragidly.
The required period of oscillation.
$T = 2 π \sqrt{\frac{μ}{k}} = 2 π \sqrt{\frac{m / 2}{k}} = π \sqrt{\frac{2 m}{k}}$

PHXI14:OSCILLATIONS

364387 A mass $M$ is suspended with a light spring. An additional mass $m$ added displaces the spring further by a distance $x$ Now, the combined mass will oscillate on the spring with period

1

T = 2 π \sqrt{\frac{m g}{x (M + m)}}

2

T = 2 π \sqrt{\frac{(M + m) x}{m g}}

3

T = \frac{π}{2} \sqrt{\frac{m g}{x (M + m)}}

4

T = 2 π \sqrt{\frac{M + m}{m g x}}

Explanation:

$M g = k x_{1} (1)$
$(M + m) g = k x_{2} (2)$
On substracting Eq. (1) from Eq. (2),
$\begin{aligned} m g = k (x_{2} - x_{1}) \\ k = \frac{m g}{x} (∵ x_{2} - x_{1} = x) \end{aligned}$
Time period $T = 2 π \sqrt{\frac{(M + m)}{k}}$
$= 2 π \sqrt{\frac{(M + m) x}{m g}}$

PHXI14:OSCILLATIONS

364388 A block of mass $M$ is suspended from a light spring of force constant $k$ . Another mass $m$ moving upwards with velocity $v$ shifts the mass $M$ and gets embeded in it. What will be the amplitude of the combined mass?

1

\frac{m v}{\sqrt{(M - m) k}}

2

\frac{m v}{k \sqrt{(M - m)}}

3

\frac{m v}{\sqrt{(M + m) k}}

4

\frac{m v}{k \sqrt{(M + m)}}

Explanation:

If $v^{'}$ are the velocities of the block of mass $M$ and $(M + m)$ while passing from the mean position when executing SHM.
Using law of conservation of linear momentum, we have
$\begin{aligned} m v & = (M + m) v^{'} \\ or v^{'} & = m v / (M + m) \end{aligned}$
Also, maximum $P E =$ maximum $K E$
$\begin{matrix} ∴ \frac{1}{2} k A^{' 2} = \frac{1}{2} (M + m) v^{' 2} \\ A^{'} = {(\frac{M + m}{k})}^{1 / 2} \times \frac{m v}{(M + m)} = \frac{m v}{\sqrt{(M + m) k}} \end{matrix}$

PHXI14:OSCILLATIONS

364389 The masses $m_{1}$ and $m_{2}$ are suspended together by a massless spring of constant $k$ . When the masses are in equilibrium, $m_{1}$ is removed without disturbing the system; the amplitude of vibration is -

1

m_{2} g / k

2

m_{1} g / k

3

\frac{(m_{2} - m_{1}) g}{k}

4

\frac{(m_{1} + m_{2}) g}{k}

Explanation:

In equilibrium $F = (m_{1} + m_{2}) g$
By removing mass $m_{1}$
$\begin{aligned} F^{'} = m_{2} g \\ F_{r} = F^{'} - F = m_{2} g - (m_{1} + m_{2}) g \\ k A = m_{1} g \Rightarrow A = \frac{m_{1} g}{k} \end{aligned}$
since the lower spring exerts force on the block only in the downward motion.

PHXI14:OSCILLATIONS

364385 An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $15 c m / \sec$ and the time $\frac{π}{5} \sec$ , find its amplitude

1 2.0

2 3.0

3 1.0

4 1.5

Explanation:

Maximum velocity, $v_{max} = a ω$
$o r v_{max} = a (\frac{2 π}{T})$
$a = \frac{v_{max} T}{2 π} = \frac{15 \times 3.14}{2 \times 3.14 \times 5} = 1.5 c m$

PHXI14:OSCILLATIONS

364386 Two identical particles each of mass $m$ are interconnected by a light spring of stiffness $k$ , the time period for small oscillation is equal to:
supporting img

1

\frac{π}{2} \sqrt{\frac{m}{k}}

2

2 π \sqrt{\frac{m}{k}}

3

π \sqrt{\frac{2 m}{k}}

4

π \sqrt{\frac{m}{2 k}}

Explanation:

The reduced mass
$μ = \frac{(m) (m)}{m + m} = \frac{m}{2}$
The given system is equivalent to a system of a particle of mass $m / 2$ connected to a spring of stiffness $k$ ragidly.
The required period of oscillation.
$T = 2 π \sqrt{\frac{μ}{k}} = 2 π \sqrt{\frac{m / 2}{k}} = π \sqrt{\frac{2 m}{k}}$

PHXI14:OSCILLATIONS

364387 A mass $M$ is suspended with a light spring. An additional mass $m$ added displaces the spring further by a distance $x$ Now, the combined mass will oscillate on the spring with period

1

T = 2 π \sqrt{\frac{m g}{x (M + m)}}

2

T = 2 π \sqrt{\frac{(M + m) x}{m g}}

3

T = \frac{π}{2} \sqrt{\frac{m g}{x (M + m)}}

4

T = 2 π \sqrt{\frac{M + m}{m g x}}

Explanation:

$M g = k x_{1} (1)$
$(M + m) g = k x_{2} (2)$
On substracting Eq. (1) from Eq. (2),
$\begin{aligned} m g = k (x_{2} - x_{1}) \\ k = \frac{m g}{x} (∵ x_{2} - x_{1} = x) \end{aligned}$
Time period $T = 2 π \sqrt{\frac{(M + m)}{k}}$
$= 2 π \sqrt{\frac{(M + m) x}{m g}}$

PHXI14:OSCILLATIONS

364388 A block of mass $M$ is suspended from a light spring of force constant $k$ . Another mass $m$ moving upwards with velocity $v$ shifts the mass $M$ and gets embeded in it. What will be the amplitude of the combined mass?

1

\frac{m v}{\sqrt{(M - m) k}}

2

\frac{m v}{k \sqrt{(M - m)}}

3

\frac{m v}{\sqrt{(M + m) k}}

4

\frac{m v}{k \sqrt{(M + m)}}

Explanation:

If $v^{'}$ are the velocities of the block of mass $M$ and $(M + m)$ while passing from the mean position when executing SHM.
Using law of conservation of linear momentum, we have
$\begin{aligned} m v & = (M + m) v^{'} \\ or v^{'} & = m v / (M + m) \end{aligned}$
Also, maximum $P E =$ maximum $K E$
$\begin{matrix} ∴ \frac{1}{2} k A^{' 2} = \frac{1}{2} (M + m) v^{' 2} \\ A^{'} = {(\frac{M + m}{k})}^{1 / 2} \times \frac{m v}{(M + m)} = \frac{m v}{\sqrt{(M + m) k}} \end{matrix}$

PHXI14:OSCILLATIONS

364389 The masses $m_{1}$ and $m_{2}$ are suspended together by a massless spring of constant $k$ . When the masses are in equilibrium, $m_{1}$ is removed without disturbing the system; the amplitude of vibration is -

1

m_{2} g / k

2

m_{1} g / k

3

\frac{(m_{2} - m_{1}) g}{k}

4

\frac{(m_{1} + m_{2}) g}{k}

Explanation:

In equilibrium $F = (m_{1} + m_{2}) g$
By removing mass $m_{1}$
$\begin{aligned} F^{'} = m_{2} g \\ F_{r} = F^{'} - F = m_{2} g - (m_{1} + m_{2}) g \\ k A = m_{1} g \Rightarrow A = \frac{m_{1} g}{k} \end{aligned}$
since the lower spring exerts force on the block only in the downward motion.

PHXI14:OSCILLATIONS

364385 An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $15 c m / \sec$ and the time $\frac{π}{5} \sec$ , find its amplitude

1 2.0

2 3.0

3 1.0

4 1.5

Explanation:

Maximum velocity, $v_{max} = a ω$
$o r v_{max} = a (\frac{2 π}{T})$
$a = \frac{v_{max} T}{2 π} = \frac{15 \times 3.14}{2 \times 3.14 \times 5} = 1.5 c m$

PHXI14:OSCILLATIONS

364386 Two identical particles each of mass $m$ are interconnected by a light spring of stiffness $k$ , the time period for small oscillation is equal to:
supporting img

1

\frac{π}{2} \sqrt{\frac{m}{k}}

2

2 π \sqrt{\frac{m}{k}}

3

π \sqrt{\frac{2 m}{k}}

4

π \sqrt{\frac{m}{2 k}}

Explanation:

The reduced mass
$μ = \frac{(m) (m)}{m + m} = \frac{m}{2}$
The given system is equivalent to a system of a particle of mass $m / 2$ connected to a spring of stiffness $k$ ragidly.
The required period of oscillation.
$T = 2 π \sqrt{\frac{μ}{k}} = 2 π \sqrt{\frac{m / 2}{k}} = π \sqrt{\frac{2 m}{k}}$

PHXI14:OSCILLATIONS

364387 A mass $M$ is suspended with a light spring. An additional mass $m$ added displaces the spring further by a distance $x$ Now, the combined mass will oscillate on the spring with period

1

T = 2 π \sqrt{\frac{m g}{x (M + m)}}

2

T = 2 π \sqrt{\frac{(M + m) x}{m g}}

3

T = \frac{π}{2} \sqrt{\frac{m g}{x (M + m)}}

4

T = 2 π \sqrt{\frac{M + m}{m g x}}

Explanation:

$M g = k x_{1} (1)$
$(M + m) g = k x_{2} (2)$
On substracting Eq. (1) from Eq. (2),
$\begin{aligned} m g = k (x_{2} - x_{1}) \\ k = \frac{m g}{x} (∵ x_{2} - x_{1} = x) \end{aligned}$
Time period $T = 2 π \sqrt{\frac{(M + m)}{k}}$
$= 2 π \sqrt{\frac{(M + m) x}{m g}}$

PHXI14:OSCILLATIONS

364388 A block of mass $M$ is suspended from a light spring of force constant $k$ . Another mass $m$ moving upwards with velocity $v$ shifts the mass $M$ and gets embeded in it. What will be the amplitude of the combined mass?

1

\frac{m v}{\sqrt{(M - m) k}}

2

\frac{m v}{k \sqrt{(M - m)}}

3

\frac{m v}{\sqrt{(M + m) k}}

4

\frac{m v}{k \sqrt{(M + m)}}

Explanation:

If $v^{'}$ are the velocities of the block of mass $M$ and $(M + m)$ while passing from the mean position when executing SHM.
Using law of conservation of linear momentum, we have
$\begin{aligned} m v & = (M + m) v^{'} \\ or v^{'} & = m v / (M + m) \end{aligned}$
Also, maximum $P E =$ maximum $K E$
$\begin{matrix} ∴ \frac{1}{2} k A^{' 2} = \frac{1}{2} (M + m) v^{' 2} \\ A^{'} = {(\frac{M + m}{k})}^{1 / 2} \times \frac{m v}{(M + m)} = \frac{m v}{\sqrt{(M + m) k}} \end{matrix}$

PHXI14:OSCILLATIONS

364389 The masses $m_{1}$ and $m_{2}$ are suspended together by a massless spring of constant $k$ . When the masses are in equilibrium, $m_{1}$ is removed without disturbing the system; the amplitude of vibration is -

1

m_{2} g / k

2

m_{1} g / k

3

\frac{(m_{2} - m_{1}) g}{k}

4

\frac{(m_{1} + m_{2}) g}{k}

Explanation:

In equilibrium $F = (m_{1} + m_{2}) g$
By removing mass $m_{1}$
$\begin{aligned} F^{'} = m_{2} g \\ F_{r} = F^{'} - F = m_{2} g - (m_{1} + m_{2}) g \\ k A = m_{1} g \Rightarrow A = \frac{m_{1} g}{k} \end{aligned}$
since the lower spring exerts force on the block only in the downward motion.

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