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A body executes 40 oscillations per minute. max speed is 36cm/s. calculate the amplitude. give the way of solution. plz,,,
Verified Answer
A body executes 40 oscillations per minute. max speed is 36cm/s. calcu...
The formula for displacement in simple harmonic motion is: 


x = A sin (
ω
t) 


The derivative would therefore give speed: 

x' = A
ω
cos (
ω
t) 


The maximum speed is when cos(
ω
t) is at the maximum or 1. So max speed is: 

v(max) = A
ω

36cm/s = A ((40 oscillations/minute) * (1 minute/second) * (2
π
/1 oscillation)) 

A = 36cm s-1 / 4.188s-1 

Amplitude A = 8.59 cm
.
This question is part of UPSC exam. View all Class 11 courses
Most Upvoted Answer
A body executes 40 oscillations per minute. max speed is 36cm/s. calcu...
Given:
- Number of oscillations per minute = 40
- Maximum speed = 36 cm/s

To find:
- Amplitude of the body's motion

Formula:
The formula to calculate the amplitude of an oscillating body is:
Amplitude (A) = Maximum speed (V_max) / Angular frequency (ω)

The angular frequency (ω) can be calculated as:
Angular frequency (ω) = 2π × Frequency (f)

Frequency (f) can be calculated as:
Frequency (f) = Number of oscillations (n) / Time

Solution:

1. Convert the given number of oscillations per minute to oscillations per second:
Number of oscillations per second = 40 / 60 = 2/3 oscillations per second

2. Calculate the time taken for one oscillation:
Time = 1 / Number of oscillations per second = 1 / (2/3) = 3/2 seconds

3. Calculate the frequency:
Frequency (f) = 1 / Time = 1 / (3/2) = 2/3 Hz

4. Calculate the angular frequency:
Angular frequency (ω) = 2π × Frequency (f) = 2π × (2/3) = 4π/3 rad/s

5. Calculate the amplitude:
Amplitude (A) = Maximum speed (V_max) / Angular frequency (ω) = 36 cm/s / (4π/3 rad/s)

6. Simplify the expression:
Amplitude (A) = 36 cm/s × 3 / (4π) = 27 / π cm ≈ 8.59 cm

Answer:
The amplitude of the body's motion is approximately 8.59 cm.
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A body executes 40 oscillations per minute. max speed is 36cm/s. calculate the amplitude. give the way of solution. plz,,,
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