First slide

Simple harmonic motion

Question

A particle executing simple harmonic motion is given by $y = 8 \sin π (10 t - \frac{1}{2})$ where y is in metre, and t is in seconds. Then the average speed of the particle during one full oscillation is

Moderate

A

80 m/s
B

0.8 m/s
C

40 m/s
D

160 m/s

Solution

Average speed of a particle executing simple harmonic motion can be written as $\frac{Total Distance}{Total time}$
In a simple harmonic motion, the total distance travelled by the particle is 4 times the amplitude. The total time taken is one full time period of oscillation.
Therefore, $Average speed= \frac{Total Distance}{Total time} = \frac{4A}{T}$
Comparing the given equation $y = 8 \sin π (10 t - \frac{1}{2})$ with standard equation, $y = A \sin (ω t + ϕ)$ , we get $A = 8 m$ and $ω = 10 π$ .
Using the relation $ω = \frac{2 π}{T}$ , we get $10 π = \frac{2 π}{T}$
i.e. $T = \frac{1}{5} s$ .
Hence $Average speed= \frac{4A}{T} = \frac{4 \times 8}{\frac{1}{5}} = 160 m / s$ .

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