Question:
A block of mass \(100\ \text{g}\) is connected to an elastic spring of spring constant \(450\ \text{N m}^{-1}\) and oscillates vertically. The block-spring system is in a viscous surrounding medium with a damping constant \(69.3\ \text{g s}^{-1}\). The time in which the amplitude of oscillations drops to half of its initial value is
\[
[\text{take } \ln 2=0.693]
\]
A block of mass \(100\ \text{g}\) is connected to an elastic spring of spring constant \(450\ \text{N m}^{-1}\) and oscillates vertically. The block-spring system is in a viscous surrounding medium with a damping constant \(69.3\ \text{g s}^{-1}\). The time in which the amplitude of oscillations drops to half of its initial value is
\[ [\text{take } \ln 2=0.693] \]
\[ [\text{take } \ln 2=0.693] \]
Show Hint
For a damped oscillator, the amplitude varies as
\[
A=A_0e^{-\frac{bt}{2m}}.
\]
When amplitude becomes half,
\[
t=\frac{2m\ln2}{b}.
\]
Updated On: Jun 26, 2026
- \(6.93\ \text{s}\)
- \(2\ \text{s}\)
- \(20\ \text{s}\)
- \(69.3\ \text{s}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Write the formula for amplitude in damped oscillations.
For a damped oscillator, amplitude decreases exponentially with time: \[ A=A_0e^{-\frac{bt}{2m}}, \] where \[ A_0=\text{initial amplitude}, \] \[ b=\text{damping constant}, \] and \[ m=\text{mass of the block}. \]
Step 2: Substitute the condition for half amplitude.
The amplitude becomes half of its initial value, so \[ A=\frac{A_0}{2}. \] Thus, \[ \frac{A_0}{2}=A_0e^{-\frac{bt}{2m}}. \] Canceling \(A_0\), \[ \frac{1}{2}=e^{-\frac{bt}{2m}}. \] Taking natural logarithm on both sides, \[ \ln\frac{1}{2}=-\frac{bt}{2m}. \] Since \[ \ln\frac{1}{2}=-\ln2, \] we get \[ \ln2=\frac{bt}{2m}. \] Therefore, \[ t=\frac{2m\ln2}{b}. \]
Step 3: Convert given quantities into consistent units.
Mass of the block is \[ m=100\ \text{g}. \] Damping constant is \[ b=69.3\ \text{g s}^{-1}. \] Since both are in gram-based units, we can use them directly in the formula.
Also, \[ \ln2=0.693. \]
Step 4: Calculate the time.
\[ t=\frac{2m\ln2}{b}. \] Substituting the values, \[ t=\frac{2(100)(0.693)}{69.3}. \] \[ t=\frac{138.6}{69.3}. \] \[ t=2\ \text{s}. \]
Step 5: Final conclusion.
Therefore, the time in which the amplitude drops to half of its initial value is \[ \boxed{2\ \text{s}} \] Hence, the correct option is \[ \boxed{(2)} \]
For a damped oscillator, amplitude decreases exponentially with time: \[ A=A_0e^{-\frac{bt}{2m}}, \] where \[ A_0=\text{initial amplitude}, \] \[ b=\text{damping constant}, \] and \[ m=\text{mass of the block}. \]
Step 2: Substitute the condition for half amplitude.
The amplitude becomes half of its initial value, so \[ A=\frac{A_0}{2}. \] Thus, \[ \frac{A_0}{2}=A_0e^{-\frac{bt}{2m}}. \] Canceling \(A_0\), \[ \frac{1}{2}=e^{-\frac{bt}{2m}}. \] Taking natural logarithm on both sides, \[ \ln\frac{1}{2}=-\frac{bt}{2m}. \] Since \[ \ln\frac{1}{2}=-\ln2, \] we get \[ \ln2=\frac{bt}{2m}. \] Therefore, \[ t=\frac{2m\ln2}{b}. \]
Step 3: Convert given quantities into consistent units.
Mass of the block is \[ m=100\ \text{g}. \] Damping constant is \[ b=69.3\ \text{g s}^{-1}. \] Since both are in gram-based units, we can use them directly in the formula.
Also, \[ \ln2=0.693. \]
Step 4: Calculate the time.
\[ t=\frac{2m\ln2}{b}. \] Substituting the values, \[ t=\frac{2(100)(0.693)}{69.3}. \] \[ t=\frac{138.6}{69.3}. \] \[ t=2\ \text{s}. \]
Step 5: Final conclusion.
Therefore, the time in which the amplitude drops to half of its initial value is \[ \boxed{2\ \text{s}} \] Hence, the correct option is \[ \boxed{(2)} \]
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