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PHXI14:OSCILLATIONS

364360 In figures $S_{1}$ and $S_{2}$ are identical springs. the oscillation frequency of the mass $m$ is $f$ . If one spring is removed, the frequency will become
supporting img

1

2 f

2

f

3

f / \sqrt{3}

4

f / \sqrt{2}

Explanation:

Effective spring constant $K_{e f f} = 2 K$ .
$f = \frac{1}{2 π} \sqrt{\frac{2 K}{m}}$
When one spring is removed, $K_{eff}$ becomes $K$ new frequency
$\begin{aligned} f^{'} = \frac{1}{2 π} \sqrt{\frac{K}{m}} \\ \frac{f^{'}}{f} = \frac{1}{\sqrt{2}} \Rightarrow f^{'} = \frac{f}{\sqrt{2}} \end{aligned}$

PHXI14:OSCILLATIONS

364361 A $1 k g$ block attached to a spring vibrates with a frequency of $1 H z$ on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an $8 k g$ block placed on the same table. So, the frequency of vibration of the $8 k g$ block is -

1

2 H z

2

\frac{1}{4} H z

3

\frac{1}{2 \sqrt{2}} H z

4

\frac{1}{2} H z

Explanation:

In first case $f = \frac{1}{2 π} \sqrt{\frac{k}{m}} = 1$
$\begin{aligned} 4 π^{2} = \frac{k}{m} \\ k = 4 π^{2} m = 1 \end{aligned}$
In second case
In parallel $k_{e q} = 2 k$
$f = \frac{1}{2 π} \sqrt{\frac{k \times 2}{8}} = \frac{1}{2} H z$

JEE - 2017

PHXI14:OSCILLATIONS

364362 Five identical springs are used in the three configuration as shown in figure. The time period of vertical oscillations in configuration (a), (b) and (c) are in the ratio
supporting img

1

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

2

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

3

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

4

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

Explanation:

$T_{a} = 2 π \sqrt{\frac{m}{k}}$
$T_{b} = 2 π \sqrt{\frac{m}{k / 2}}$
$T_{c} = 2 π \sqrt{\frac{m}{2 k}}$

PHXI14:OSCILLATIONS

364363 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 the mass $M$ is displaced in the horizontal direction, then find the frequency of the system.
supporting img

1

\frac{1}{2 π} \sqrt{\frac{k}{4 M}}

2

\frac{1}{2 π} \sqrt{\frac{4 k}{M}}

3

\frac{1}{2 π} \sqrt{\frac{k}{7 M}}

4

\frac{1}{2 π} \sqrt{\frac{7 k}{M}}

Explanation:

Spring on the left of the block are in series, hence there equivalent is
$k_{1} = \frac{(2 k) (2 k)}{2 k + 2 k} = k$
Springs on the right of the block are in parallel, hence there equivalent is
$k_{2} = k + 2 k = 3 k$
Now again both $K_{1}$ and $K_{2}$ are in parallel
$∴ k_{eq.} = k_{1} + k_{2} = k + 3 k = 4 k$
Hence, frequency is
$f = \frac{1}{2 π} \sqrt{\frac{k_{e q}}{M}} = \frac{1}{2 π} \sqrt{\frac{4 k}{M}}$

PHXI14:OSCILLATIONS

364360 In figures $S_{1}$ and $S_{2}$ are identical springs. the oscillation frequency of the mass $m$ is $f$ . If one spring is removed, the frequency will become
supporting img

1

2 f

2

f

3

f / \sqrt{3}

4

f / \sqrt{2}

Explanation:

Effective spring constant $K_{e f f} = 2 K$ .
$f = \frac{1}{2 π} \sqrt{\frac{2 K}{m}}$
When one spring is removed, $K_{eff}$ becomes $K$ new frequency
$\begin{aligned} f^{'} = \frac{1}{2 π} \sqrt{\frac{K}{m}} \\ \frac{f^{'}}{f} = \frac{1}{\sqrt{2}} \Rightarrow f^{'} = \frac{f}{\sqrt{2}} \end{aligned}$

PHXI14:OSCILLATIONS

364361 A $1 k g$ block attached to a spring vibrates with a frequency of $1 H z$ on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an $8 k g$ block placed on the same table. So, the frequency of vibration of the $8 k g$ block is -

1

2 H z

2

\frac{1}{4} H z

3

\frac{1}{2 \sqrt{2}} H z

4

\frac{1}{2} H z

Explanation:

In first case $f = \frac{1}{2 π} \sqrt{\frac{k}{m}} = 1$
$\begin{aligned} 4 π^{2} = \frac{k}{m} \\ k = 4 π^{2} m = 1 \end{aligned}$
In second case
In parallel $k_{e q} = 2 k$
$f = \frac{1}{2 π} \sqrt{\frac{k \times 2}{8}} = \frac{1}{2} H z$

JEE - 2017

PHXI14:OSCILLATIONS

364362 Five identical springs are used in the three configuration as shown in figure. The time period of vertical oscillations in configuration (a), (b) and (c) are in the ratio
supporting img

1

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

2

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

3

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

4

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

Explanation:

$T_{a} = 2 π \sqrt{\frac{m}{k}}$
$T_{b} = 2 π \sqrt{\frac{m}{k / 2}}$
$T_{c} = 2 π \sqrt{\frac{m}{2 k}}$

PHXI14:OSCILLATIONS

364363 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 the mass $M$ is displaced in the horizontal direction, then find the frequency of the system.
supporting img

1

\frac{1}{2 π} \sqrt{\frac{k}{4 M}}

2

\frac{1}{2 π} \sqrt{\frac{4 k}{M}}

3

\frac{1}{2 π} \sqrt{\frac{k}{7 M}}

4

\frac{1}{2 π} \sqrt{\frac{7 k}{M}}

Explanation:

Spring on the left of the block are in series, hence there equivalent is
$k_{1} = \frac{(2 k) (2 k)}{2 k + 2 k} = k$
Springs on the right of the block are in parallel, hence there equivalent is
$k_{2} = k + 2 k = 3 k$
Now again both $K_{1}$ and $K_{2}$ are in parallel
$∴ k_{eq.} = k_{1} + k_{2} = k + 3 k = 4 k$
Hence, frequency is
$f = \frac{1}{2 π} \sqrt{\frac{k_{e q}}{M}} = \frac{1}{2 π} \sqrt{\frac{4 k}{M}}$

PHXI14:OSCILLATIONS

364360 In figures $S_{1}$ and $S_{2}$ are identical springs. the oscillation frequency of the mass $m$ is $f$ . If one spring is removed, the frequency will become
supporting img

1

2 f

2

f

3

f / \sqrt{3}

4

f / \sqrt{2}

Explanation:

Effective spring constant $K_{e f f} = 2 K$ .
$f = \frac{1}{2 π} \sqrt{\frac{2 K}{m}}$
When one spring is removed, $K_{eff}$ becomes $K$ new frequency
$\begin{aligned} f^{'} = \frac{1}{2 π} \sqrt{\frac{K}{m}} \\ \frac{f^{'}}{f} = \frac{1}{\sqrt{2}} \Rightarrow f^{'} = \frac{f}{\sqrt{2}} \end{aligned}$

PHXI14:OSCILLATIONS

364361 A $1 k g$ block attached to a spring vibrates with a frequency of $1 H z$ on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an $8 k g$ block placed on the same table. So, the frequency of vibration of the $8 k g$ block is -

1

2 H z

2

\frac{1}{4} H z

3

\frac{1}{2 \sqrt{2}} H z

4

\frac{1}{2} H z

Explanation:

In first case $f = \frac{1}{2 π} \sqrt{\frac{k}{m}} = 1$
$\begin{aligned} 4 π^{2} = \frac{k}{m} \\ k = 4 π^{2} m = 1 \end{aligned}$
In second case
In parallel $k_{e q} = 2 k$
$f = \frac{1}{2 π} \sqrt{\frac{k \times 2}{8}} = \frac{1}{2} H z$

JEE - 2017

PHXI14:OSCILLATIONS

364362 Five identical springs are used in the three configuration as shown in figure. The time period of vertical oscillations in configuration (a), (b) and (c) are in the ratio
supporting img

1

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

2

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

3

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

4

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

Explanation:

$T_{a} = 2 π \sqrt{\frac{m}{k}}$
$T_{b} = 2 π \sqrt{\frac{m}{k / 2}}$
$T_{c} = 2 π \sqrt{\frac{m}{2 k}}$

PHXI14:OSCILLATIONS

364363 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 the mass $M$ is displaced in the horizontal direction, then find the frequency of the system.
supporting img

1

\frac{1}{2 π} \sqrt{\frac{k}{4 M}}

2

\frac{1}{2 π} \sqrt{\frac{4 k}{M}}

3

\frac{1}{2 π} \sqrt{\frac{k}{7 M}}

4

\frac{1}{2 π} \sqrt{\frac{7 k}{M}}

Explanation:

Spring on the left of the block are in series, hence there equivalent is
$k_{1} = \frac{(2 k) (2 k)}{2 k + 2 k} = k$
Springs on the right of the block are in parallel, hence there equivalent is
$k_{2} = k + 2 k = 3 k$
Now again both $K_{1}$ and $K_{2}$ are in parallel
$∴ k_{eq.} = k_{1} + k_{2} = k + 3 k = 4 k$
Hence, frequency is
$f = \frac{1}{2 π} \sqrt{\frac{k_{e q}}{M}} = \frac{1}{2 π} \sqrt{\frac{4 k}{M}}$

PHXI14:OSCILLATIONS

364360 In figures $S_{1}$ and $S_{2}$ are identical springs. the oscillation frequency of the mass $m$ is $f$ . If one spring is removed, the frequency will become
supporting img

1

2 f

2

f

3

f / \sqrt{3}

4

f / \sqrt{2}

Explanation:

Effective spring constant $K_{e f f} = 2 K$ .
$f = \frac{1}{2 π} \sqrt{\frac{2 K}{m}}$
When one spring is removed, $K_{eff}$ becomes $K$ new frequency
$\begin{aligned} f^{'} = \frac{1}{2 π} \sqrt{\frac{K}{m}} \\ \frac{f^{'}}{f} = \frac{1}{\sqrt{2}} \Rightarrow f^{'} = \frac{f}{\sqrt{2}} \end{aligned}$

PHXI14:OSCILLATIONS

364361 A $1 k g$ block attached to a spring vibrates with a frequency of $1 H z$ on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an $8 k g$ block placed on the same table. So, the frequency of vibration of the $8 k g$ block is -

1

2 H z

2

\frac{1}{4} H z

3

\frac{1}{2 \sqrt{2}} H z

4

\frac{1}{2} H z

Explanation:

In first case $f = \frac{1}{2 π} \sqrt{\frac{k}{m}} = 1$
$\begin{aligned} 4 π^{2} = \frac{k}{m} \\ k = 4 π^{2} m = 1 \end{aligned}$
In second case
In parallel $k_{e q} = 2 k$
$f = \frac{1}{2 π} \sqrt{\frac{k \times 2}{8}} = \frac{1}{2} H z$

JEE - 2017

PHXI14:OSCILLATIONS

364362 Five identical springs are used in the three configuration as shown in figure. The time period of vertical oscillations in configuration (a), (b) and (c) are in the ratio
supporting img

1

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

2

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

3

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

4

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

Explanation:

$T_{a} = 2 π \sqrt{\frac{m}{k}}$
$T_{b} = 2 π \sqrt{\frac{m}{k / 2}}$
$T_{c} = 2 π \sqrt{\frac{m}{2 k}}$

PHXI14:OSCILLATIONS

364363 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 the mass $M$ is displaced in the horizontal direction, then find the frequency of the system.
supporting img

1

\frac{1}{2 π} \sqrt{\frac{k}{4 M}}

2

\frac{1}{2 π} \sqrt{\frac{4 k}{M}}

3

\frac{1}{2 π} \sqrt{\frac{k}{7 M}}

4

\frac{1}{2 π} \sqrt{\frac{7 k}{M}}

Explanation:

Spring on the left of the block are in series, hence there equivalent is
$k_{1} = \frac{(2 k) (2 k)}{2 k + 2 k} = k$
Springs on the right of the block are in parallel, hence there equivalent is
$k_{2} = k + 2 k = 3 k$
Now again both $K_{1}$ and $K_{2}$ are in parallel
$∴ k_{eq.} = k_{1} + k_{2} = k + 3 k = 4 k$
Hence, frequency is
$f = \frac{1}{2 π} \sqrt{\frac{k_{e q}}{M}} = \frac{1}{2 π} \sqrt{\frac{4 k}{M}}$