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A mass is suspended from a vertica sprin...

A mass is suspended from a vertica spring which is executing SHM of frequency 5 Hz.
The spring is unstretched at the highest point of oscillation. Maximum speed of the mass is (take, acceleration due to gravity, `g=10m//s^(2)` )

A

`2 pi ms^(-1)`

B

` pi ms^(-1)`

C

` 1/(2pi) ms^(-1)`

D

` 1/(pi) ms^(-1)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the maximum speed of a mass suspended from a vertical spring executing simple harmonic motion (SHM) with a frequency of 5 Hz, we can follow these steps: ### Step 1: Understand the relationship between frequency, spring constant, and mass The frequency \( f \) of a mass-spring system in SHM is given by the formula: \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \] where \( k \) is the spring constant and \( m \) is the mass. ### Step 2: Rearrange the formula to find \( \frac{k}{m} \) From the frequency equation, we can rearrange it to find \( \frac{k}{m} \): \[ \sqrt{\frac{k}{m}} = 2\pi f \] Squaring both sides gives: \[ \frac{k}{m} = (2\pi f)^2 \] ### Step 3: Substitute the given frequency Substituting \( f = 5 \, \text{Hz} \): \[ \frac{k}{m} = (2\pi \cdot 5)^2 = (10\pi)^2 = 100\pi^2 \] ### Step 4: Find the amplitude of the oscillation At the highest point of oscillation, the spring is unstretched, and the maximum extension (amplitude \( A \)) can be expressed as: \[ A = \frac{mg}{k} \] Since \( k = \frac{100\pi^2 m}{1} \) from the previous step, we can substitute this into the amplitude equation: \[ A = \frac{mg}{\frac{100\pi^2 m}{1}} = \frac{g}{100\pi^2} \] ### Step 5: Calculate the maximum speed The maximum speed \( V_{\text{max}} \) in SHM is given by: \[ V_{\text{max}} = \omega A \] where \( \omega = 2\pi f \). We already know \( f = 5 \, \text{Hz} \), so: \[ \omega = 2\pi \cdot 5 = 10\pi \] Now substituting for \( A \): \[ V_{\text{max}} = 10\pi \cdot \frac{g}{100\pi^2} \] Substituting \( g = 10 \, \text{m/s}^2 \): \[ V_{\text{max}} = 10\pi \cdot \frac{10}{100\pi^2} = \frac{100\pi}{100\pi^2} = \frac{1}{\pi} \] ### Final Answer Thus, the maximum speed of the mass is: \[ V_{\text{max}} = \frac{1}{\pi} \, \text{m/s} \] ---
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