QBManager

PHXI14:OSCILLATIONS

364364 A block of mass $m$ is tied to one end of a spring which passes over a smooth fixed pulley $A$ and under a light smooth movable pulley $B$ . The other end of the string is attached to the lower end of a spring of spring constant $K_{2}$ . Find the period of small oscillation of mass $m$ about its equilibrium position.
(Take $m = π^{2} k g$ , $K_{2} = 4 K_{1}$ and $K_{1} = 17 N / m$ .
supporting img

1

1 s

2

3 s

3

7 s

4

9 s

Explanation:

Net increment in the tension in string connecting the block will provide acceleration to the block. We can write
$Δ T = m a (1)$
When the block $m$ is displaced by a distance $x$ beyond equilibrium position, the addition stretch of springs 1 and 2 are $x_{1}$ and $x_{2}$ , respectively.We can write
$x_{1} = \frac{x - x_{2}}{2} \Rightarrow x = 2 x_{1} + x_{2} (2)$
As $Δ T = K_{2} x_{2}$ and $2 Δ T = K_{1} x_{1}$
$2 (K_{2} x_{2}) = K_{1} x_{1} \Rightarrow x_{1} = \frac{2 K_{2} x_{2}}{K_{1}} (3)$
From Eqs. (2) and (3),
$x = 2 [\frac{2 K_{2} x_{2}}{K_{1}}] + x_{2} = \frac{(4 K_{2} + K_{1})}{K_{1}} x_{2}$
$\Rightarrow x_{2} = \frac{K_{1} x}{(K_{1} + 4 K_{2})}$
$a n d \vec{a} = \frac{- K_{1} K_{2}}{(K_{1} + 4 K_{2}) m} \cdot x$
$\Rightarrow \vec{a} = \frac{- K_{1} K_{2}}{(K_{1} + 4 K_{2}) m} \cdot x$
$ω^{2} = \frac{K_{1} K_{2}}{(K_{1} + 4 K_{2}) m}$
Hence $R = \frac{80}{23 π^{2}} m$
After substituting the values, we get $T = 1 s$ .
supporting img

PHXI14:OSCILLATIONS

364365 If each spring in figure has force constant of $10 \frac{N}{m}$ , and attach mass is $10 k g$ , then find the frequency of oscillation.
supporting img

1

π H z

2

\frac{1}{π} H z

3

2 H z

4

\frac{2}{π} H z

Explanation:

$T = 2 π \sqrt{\frac{m}{k_{e f f}}}$
All springs are connected in parallel.
$\begin{matrix} k_{eff.} = 40 N / m \\ T = 2 π \sqrt{\frac{10}{40}} = π \\ \Rightarrow f = \frac{1}{π} H z \end{matrix}$

PHXI14:OSCILLATIONS

364366 A uniform stick of mass $M$ and length $L$ is pivoted at its centre. Its ends are tied to two springs each of force constant $K$ . In the position shown figure, the strings are in their natural length. When the string is displaced through a small angle $θ$ and released. The stick:
supporting img

1 Executes periodic motion which is not simple harmonic

2 Executes non-periodic motion

3 Executes S.H.M. of frequency

\frac{1}{2 π} \sqrt{\frac{6 K}{M}}

4 Executes S.H.M. of frequency

\frac{1}{2 π} \sqrt{\frac{K}{2 M}}

Explanation:

If stick is rotated through a small angle $θ$ , the spring is stretched by a distance $\frac{L}{2} θ$ . Therefore restoring force is spring $= K (\frac{L}{2} θ)$ . Each spring causes a torque $τ (= F d) = (\frac{K L θ}{2}) \frac{L}{2}$ in the same direction.
Therefore equation of rotational motion is
$\begin{aligned} τ = I α \\ - 2 (\frac{K L θ}{2}) \frac{L}{2} = I α with I = \frac{M L^{2}}{12} \end{aligned}$
$α = - (\frac{6 K}{M}) θ$
Standard equation of angular S.H.M. is
$\begin{aligned} α = - ω^{2} θ \\ ω^{2} = \frac{6 K}{M} \end{aligned}$
frequency $f = \frac{ω}{2 π} = \frac{1}{2 π} \sqrt{\frac{6 K}{M}}$

PHXI14:OSCILLATIONS

364367 Two blocks each of mass $m = 1 k g$ are connected to the spring of spring constant $k = 2$ units (as in fig.) If the blocks are displaced slightly in opposite directions and released, they will execute SHM. What is the time period of oscillation ?
supporting img

1

3.14 s

2

1.27 s

3

6.13 s

4

4.25 s

Explanation:

Reduced mass of the system,
$μ = \frac{m (m)}{m + m} = \frac{m}{2}$
$∴$ Time period $T = 2 π \sqrt{\frac{μ}{k}}$
$= 2 π \sqrt{\frac{1}{2 \times 2}}$
$= π$
$∴ T = 3.14 s$

PHXI14:OSCILLATIONS

364364 A block of mass $m$ is tied to one end of a spring which passes over a smooth fixed pulley $A$ and under a light smooth movable pulley $B$ . The other end of the string is attached to the lower end of a spring of spring constant $K_{2}$ . Find the period of small oscillation of mass $m$ about its equilibrium position.
(Take $m = π^{2} k g$ , $K_{2} = 4 K_{1}$ and $K_{1} = 17 N / m$ .
supporting img

1

1 s

2

3 s

3

7 s

4

9 s

Explanation:

Net increment in the tension in string connecting the block will provide acceleration to the block. We can write
$Δ T = m a (1)$
When the block $m$ is displaced by a distance $x$ beyond equilibrium position, the addition stretch of springs 1 and 2 are $x_{1}$ and $x_{2}$ , respectively.We can write
$x_{1} = \frac{x - x_{2}}{2} \Rightarrow x = 2 x_{1} + x_{2} (2)$
As $Δ T = K_{2} x_{2}$ and $2 Δ T = K_{1} x_{1}$
$2 (K_{2} x_{2}) = K_{1} x_{1} \Rightarrow x_{1} = \frac{2 K_{2} x_{2}}{K_{1}} (3)$
From Eqs. (2) and (3),
$x = 2 [\frac{2 K_{2} x_{2}}{K_{1}}] + x_{2} = \frac{(4 K_{2} + K_{1})}{K_{1}} x_{2}$
$\Rightarrow x_{2} = \frac{K_{1} x}{(K_{1} + 4 K_{2})}$
$a n d \vec{a} = \frac{- K_{1} K_{2}}{(K_{1} + 4 K_{2}) m} \cdot x$
$\Rightarrow \vec{a} = \frac{- K_{1} K_{2}}{(K_{1} + 4 K_{2}) m} \cdot x$
$ω^{2} = \frac{K_{1} K_{2}}{(K_{1} + 4 K_{2}) m}$
Hence $R = \frac{80}{23 π^{2}} m$
After substituting the values, we get $T = 1 s$ .
supporting img

PHXI14:OSCILLATIONS

364365 If each spring in figure has force constant of $10 \frac{N}{m}$ , and attach mass is $10 k g$ , then find the frequency of oscillation.
supporting img

1

π H z

2

\frac{1}{π} H z

3

2 H z

4

\frac{2}{π} H z

Explanation:

$T = 2 π \sqrt{\frac{m}{k_{e f f}}}$
All springs are connected in parallel.
$\begin{matrix} k_{eff.} = 40 N / m \\ T = 2 π \sqrt{\frac{10}{40}} = π \\ \Rightarrow f = \frac{1}{π} H z \end{matrix}$

PHXI14:OSCILLATIONS

364366 A uniform stick of mass $M$ and length $L$ is pivoted at its centre. Its ends are tied to two springs each of force constant $K$ . In the position shown figure, the strings are in their natural length. When the string is displaced through a small angle $θ$ and released. The stick:
supporting img

1 Executes periodic motion which is not simple harmonic

2 Executes non-periodic motion

3 Executes S.H.M. of frequency

\frac{1}{2 π} \sqrt{\frac{6 K}{M}}

4 Executes S.H.M. of frequency

\frac{1}{2 π} \sqrt{\frac{K}{2 M}}

Explanation:

If stick is rotated through a small angle $θ$ , the spring is stretched by a distance $\frac{L}{2} θ$ . Therefore restoring force is spring $= K (\frac{L}{2} θ)$ . Each spring causes a torque $τ (= F d) = (\frac{K L θ}{2}) \frac{L}{2}$ in the same direction.
Therefore equation of rotational motion is
$\begin{aligned} τ = I α \\ - 2 (\frac{K L θ}{2}) \frac{L}{2} = I α with I = \frac{M L^{2}}{12} \end{aligned}$
$α = - (\frac{6 K}{M}) θ$
Standard equation of angular S.H.M. is
$\begin{aligned} α = - ω^{2} θ \\ ω^{2} = \frac{6 K}{M} \end{aligned}$
frequency $f = \frac{ω}{2 π} = \frac{1}{2 π} \sqrt{\frac{6 K}{M}}$

PHXI14:OSCILLATIONS

364367 Two blocks each of mass $m = 1 k g$ are connected to the spring of spring constant $k = 2$ units (as in fig.) If the blocks are displaced slightly in opposite directions and released, they will execute SHM. What is the time period of oscillation ?
supporting img

1

3.14 s

2

1.27 s

3

6.13 s

4

4.25 s

Explanation:

Reduced mass of the system,
$μ = \frac{m (m)}{m + m} = \frac{m}{2}$
$∴$ Time period $T = 2 π \sqrt{\frac{μ}{k}}$
$= 2 π \sqrt{\frac{1}{2 \times 2}}$
$= π$
$∴ T = 3.14 s$

PHXI14:OSCILLATIONS

364364 A block of mass $m$ is tied to one end of a spring which passes over a smooth fixed pulley $A$ and under a light smooth movable pulley $B$ . The other end of the string is attached to the lower end of a spring of spring constant $K_{2}$ . Find the period of small oscillation of mass $m$ about its equilibrium position.
(Take $m = π^{2} k g$ , $K_{2} = 4 K_{1}$ and $K_{1} = 17 N / m$ .
supporting img

1

1 s

2

3 s

3

7 s

4

9 s

Explanation:

Net increment in the tension in string connecting the block will provide acceleration to the block. We can write
$Δ T = m a (1)$
When the block $m$ is displaced by a distance $x$ beyond equilibrium position, the addition stretch of springs 1 and 2 are $x_{1}$ and $x_{2}$ , respectively.We can write
$x_{1} = \frac{x - x_{2}}{2} \Rightarrow x = 2 x_{1} + x_{2} (2)$
As $Δ T = K_{2} x_{2}$ and $2 Δ T = K_{1} x_{1}$
$2 (K_{2} x_{2}) = K_{1} x_{1} \Rightarrow x_{1} = \frac{2 K_{2} x_{2}}{K_{1}} (3)$
From Eqs. (2) and (3),
$x = 2 [\frac{2 K_{2} x_{2}}{K_{1}}] + x_{2} = \frac{(4 K_{2} + K_{1})}{K_{1}} x_{2}$
$\Rightarrow x_{2} = \frac{K_{1} x}{(K_{1} + 4 K_{2})}$
$a n d \vec{a} = \frac{- K_{1} K_{2}}{(K_{1} + 4 K_{2}) m} \cdot x$
$\Rightarrow \vec{a} = \frac{- K_{1} K_{2}}{(K_{1} + 4 K_{2}) m} \cdot x$
$ω^{2} = \frac{K_{1} K_{2}}{(K_{1} + 4 K_{2}) m}$
Hence $R = \frac{80}{23 π^{2}} m$
After substituting the values, we get $T = 1 s$ .
supporting img

PHXI14:OSCILLATIONS

364365 If each spring in figure has force constant of $10 \frac{N}{m}$ , and attach mass is $10 k g$ , then find the frequency of oscillation.
supporting img

1

π H z

2

\frac{1}{π} H z

3

2 H z

4

\frac{2}{π} H z

Explanation:

$T = 2 π \sqrt{\frac{m}{k_{e f f}}}$
All springs are connected in parallel.
$\begin{matrix} k_{eff.} = 40 N / m \\ T = 2 π \sqrt{\frac{10}{40}} = π \\ \Rightarrow f = \frac{1}{π} H z \end{matrix}$

PHXI14:OSCILLATIONS

364366 A uniform stick of mass $M$ and length $L$ is pivoted at its centre. Its ends are tied to two springs each of force constant $K$ . In the position shown figure, the strings are in their natural length. When the string is displaced through a small angle $θ$ and released. The stick:
supporting img

1 Executes periodic motion which is not simple harmonic

2 Executes non-periodic motion

3 Executes S.H.M. of frequency

\frac{1}{2 π} \sqrt{\frac{6 K}{M}}

4 Executes S.H.M. of frequency

\frac{1}{2 π} \sqrt{\frac{K}{2 M}}

Explanation:

If stick is rotated through a small angle $θ$ , the spring is stretched by a distance $\frac{L}{2} θ$ . Therefore restoring force is spring $= K (\frac{L}{2} θ)$ . Each spring causes a torque $τ (= F d) = (\frac{K L θ}{2}) \frac{L}{2}$ in the same direction.
Therefore equation of rotational motion is
$\begin{aligned} τ = I α \\ - 2 (\frac{K L θ}{2}) \frac{L}{2} = I α with I = \frac{M L^{2}}{12} \end{aligned}$
$α = - (\frac{6 K}{M}) θ$
Standard equation of angular S.H.M. is
$\begin{aligned} α = - ω^{2} θ \\ ω^{2} = \frac{6 K}{M} \end{aligned}$
frequency $f = \frac{ω}{2 π} = \frac{1}{2 π} \sqrt{\frac{6 K}{M}}$

PHXI14:OSCILLATIONS

364367 Two blocks each of mass $m = 1 k g$ are connected to the spring of spring constant $k = 2$ units (as in fig.) If the blocks are displaced slightly in opposite directions and released, they will execute SHM. What is the time period of oscillation ?
supporting img

1

3.14 s

2

1.27 s

3

6.13 s

4

4.25 s

Explanation:

Reduced mass of the system,
$μ = \frac{m (m)}{m + m} = \frac{m}{2}$
$∴$ Time period $T = 2 π \sqrt{\frac{μ}{k}}$
$= 2 π \sqrt{\frac{1}{2 \times 2}}$
$= π$
$∴ T = 3.14 s$

PHXI14:OSCILLATIONS

364364 A block of mass $m$ is tied to one end of a spring which passes over a smooth fixed pulley $A$ and under a light smooth movable pulley $B$ . The other end of the string is attached to the lower end of a spring of spring constant $K_{2}$ . Find the period of small oscillation of mass $m$ about its equilibrium position.
(Take $m = π^{2} k g$ , $K_{2} = 4 K_{1}$ and $K_{1} = 17 N / m$ .
supporting img

1

1 s

2

3 s

3

7 s

4

9 s

Explanation:

Net increment in the tension in string connecting the block will provide acceleration to the block. We can write
$Δ T = m a (1)$
When the block $m$ is displaced by a distance $x$ beyond equilibrium position, the addition stretch of springs 1 and 2 are $x_{1}$ and $x_{2}$ , respectively.We can write
$x_{1} = \frac{x - x_{2}}{2} \Rightarrow x = 2 x_{1} + x_{2} (2)$
As $Δ T = K_{2} x_{2}$ and $2 Δ T = K_{1} x_{1}$
$2 (K_{2} x_{2}) = K_{1} x_{1} \Rightarrow x_{1} = \frac{2 K_{2} x_{2}}{K_{1}} (3)$
From Eqs. (2) and (3),
$x = 2 [\frac{2 K_{2} x_{2}}{K_{1}}] + x_{2} = \frac{(4 K_{2} + K_{1})}{K_{1}} x_{2}$
$\Rightarrow x_{2} = \frac{K_{1} x}{(K_{1} + 4 K_{2})}$
$a n d \vec{a} = \frac{- K_{1} K_{2}}{(K_{1} + 4 K_{2}) m} \cdot x$
$\Rightarrow \vec{a} = \frac{- K_{1} K_{2}}{(K_{1} + 4 K_{2}) m} \cdot x$
$ω^{2} = \frac{K_{1} K_{2}}{(K_{1} + 4 K_{2}) m}$
Hence $R = \frac{80}{23 π^{2}} m$
After substituting the values, we get $T = 1 s$ .
supporting img

PHXI14:OSCILLATIONS

364365 If each spring in figure has force constant of $10 \frac{N}{m}$ , and attach mass is $10 k g$ , then find the frequency of oscillation.
supporting img

1

π H z

2

\frac{1}{π} H z

3

2 H z

4

\frac{2}{π} H z

Explanation:

$T = 2 π \sqrt{\frac{m}{k_{e f f}}}$
All springs are connected in parallel.
$\begin{matrix} k_{eff.} = 40 N / m \\ T = 2 π \sqrt{\frac{10}{40}} = π \\ \Rightarrow f = \frac{1}{π} H z \end{matrix}$

PHXI14:OSCILLATIONS

364366 A uniform stick of mass $M$ and length $L$ is pivoted at its centre. Its ends are tied to two springs each of force constant $K$ . In the position shown figure, the strings are in their natural length. When the string is displaced through a small angle $θ$ and released. The stick:
supporting img

1 Executes periodic motion which is not simple harmonic

2 Executes non-periodic motion

3 Executes S.H.M. of frequency

\frac{1}{2 π} \sqrt{\frac{6 K}{M}}

4 Executes S.H.M. of frequency

\frac{1}{2 π} \sqrt{\frac{K}{2 M}}

Explanation:

If stick is rotated through a small angle $θ$ , the spring is stretched by a distance $\frac{L}{2} θ$ . Therefore restoring force is spring $= K (\frac{L}{2} θ)$ . Each spring causes a torque $τ (= F d) = (\frac{K L θ}{2}) \frac{L}{2}$ in the same direction.
Therefore equation of rotational motion is
$\begin{aligned} τ = I α \\ - 2 (\frac{K L θ}{2}) \frac{L}{2} = I α with I = \frac{M L^{2}}{12} \end{aligned}$
$α = - (\frac{6 K}{M}) θ$
Standard equation of angular S.H.M. is
$\begin{aligned} α = - ω^{2} θ \\ ω^{2} = \frac{6 K}{M} \end{aligned}$
frequency $f = \frac{ω}{2 π} = \frac{1}{2 π} \sqrt{\frac{6 K}{M}}$

PHXI14:OSCILLATIONS

364367 Two blocks each of mass $m = 1 k g$ are connected to the spring of spring constant $k = 2$ units (as in fig.) If the blocks are displaced slightly in opposite directions and released, they will execute SHM. What is the time period of oscillation ?
supporting img

1

3.14 s

2

1.27 s

3

6.13 s

4

4.25 s

Explanation:

Reduced mass of the system,
$μ = \frac{m (m)}{m + m} = \frac{m}{2}$
$∴$ Time period $T = 2 π \sqrt{\frac{μ}{k}}$
$= 2 π \sqrt{\frac{1}{2 \times 2}}$
$= π$
$∴ T = 3.14 s$

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: