Spring Block System

Practice Questions:

Question 1: moderate

One end of an ideal spring is fixed with a wall and the other end is fixed with a block of mass 1 kg. Force constant of spring is 100 N/m and block is performing S.H.M. with amplitude 3 cm. When the block is at left extreme position, another block of mass 3 kg moving directly towards 1 kg block with velocity 80/3 cm/s collides and gets stuck to it. The amplitude of oscillation of the combined body is :

1. 3 cm

2. 4 cm

3. 5 cm

4. 6 cm

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Question 2: moderate

A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. Then the ratio of m/M is :

1. 3/5

2. 25/9

3. 16/9

4. 5/3

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1. Initial Time Period:
\[
T = 2\pi \sqrt{\frac{M}{k}}
\]

2. New Time Period:
\[
\frac{5T}{3} = 2\pi \sqrt{\frac{M + m}{k}}
\]

3. Divide the New Time Period by the Initial Time Period:

\[
\frac{\frac{5T}{3}}{T} = \sqrt{\frac{M + m}{M}}
\]

\[
\frac{5}{3} = \sqrt{\frac{M + m}{M}}
\]

4. Square Both Sides:

\[
\frac{25}{9} = \frac{M + m}{M}
\]

5. Solve for \( \frac{m}{M} \):

\[
\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}
\]

Answer:

\[
\frac{m}{M} = \frac{16}{9}
\]

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Question 3: moderate

Four massless springs whose force constants are 2k, 2k, k and 2k 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 the frequency of oscillation of the system is :

1. \[ \frac{1}{2\pi} \sqrt{\frac{k}{4M}}\]

2. \[ \frac{1}{2\pi} \sqrt{\frac{4k}{M}}\]

3. \[ \frac{1}{2\pi} \sqrt{\frac{k}{7M}}\]

4. \[ \frac{1}{2\pi} \sqrt{\frac{7k}{M}}\]

View Answer

1. Left Side:
- Two springs with spring constants \( 2k \) and \( 2k \) are in series.
- The combined spring constant \( k_{\text{left}} \) for these two springs in series is:
\[
\frac{1}{k_{\text{left}}} = \frac{1}{2k} + \frac{1}{2k} = \frac{1}{k}
\]
\[
k_{\text{left}} = k
\]

2. Right Side:
- Two springs with spring constants \( k \) and \( 2k \) are in parallel.
- The combined spring constant \( k_{\text{right}} \) for these two springs in parallel is:
\[
k_{\text{right}} = k + 2k = 3k
\]

3. Combine Left and Right Sides:
- Since \( k_{\text{left}} \) and \( k_{\text{right}} \) are in parallel, the equivalent spring constant \( k_{\text{eq}} \) is:
\[
k_{\text{eq}} = k_{\text{left}} + k_{\text{right}} = k + 3k = 4k
\]

Step 2: Calculate the Frequency of Oscillation

The frequency \( f \) is given by:

\[
f = \frac{1}{2\pi} \sqrt{\frac{k_{\text{eq}}}{M}}
\]

Substitute \( k_{\text{eq}} = 4k \):

\[
f = \frac{1}{2\pi} \sqrt{\frac{4k}{M}}
\]

Final Answer

\[
f = \frac{1}{2\pi} \sqrt{\frac{4k}{M}}
\]

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Question 4: moderate

A force of 6.4 N stretches a vertical spring by 0.1 m. The mass that must be suspended from the spring so that it oscillates with a time period of π/4 second :

1. π/4 kg

2. 4/π kg

3. 1 kg

4. 10 kg

View Answer

Given:

- Force \( F = 6.4 \, \text{N} \)
- Extension \( x = 0.1 \, \text{m} \)
- Time period \( T = \frac{\pi}{4} \, \text{s} \)

1. Find the spring constant \( k \):

\[
k = \frac{F}{x} = \frac{6.4}{0.1} = 64 \, \text{N/m}
\]

2. Use the formula for the time period of a mass-spring system:

\[
T = 2\pi \sqrt{\frac{m}{k}}
\]

Substitute \( T = \frac{\pi}{4} \) and \( k = 64 \):

\[
\frac{\pi}{4} = 2\pi \sqrt{\frac{m}{64}}
\]

3. Solve for \( m \):

\[
\frac{1}{8} = \sqrt{\frac{m}{64}}
\]

\[
\frac{1}{64} = \frac{m}{64}
\]

\[
m = 1 \, \text{kg}
\]

Answer: \( m = 1 \, \text{kg} \)

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