First slide

Simple hormonic motion

Question

Figure shows the displacement –time graphs of two simple harmonic motions I and II from the graph it follows that

Easy

A

Curve I has same frequency as that of curve –II
B

Curve I has frequency twice that of curve-II
C

Curve I has frequency half that of curve-II
D

Curve I has frequency four times that of curve-II

Solution

For curve - I, cycle time
T₁ = (5-1)sec = 4sec
For curve- II, cycle time
T₂ = (10-2)sec = 8 sec
$\large \therefore \,{n_1}\, = \,\frac{1}{{{T_1}}} = \,0.25Hz$
and $\large {n_2}\, = \,\frac{1}{{{T_2}}}\, = \,0.125Hz$

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