The motion of a mass on a spring, with spring constant $K$ is as shown in figure.
The equation of motion is given by, $x (t) = A \sin ω t +$ $B \cos ω t$ with $ω = \sqrt{\frac{K}{m}}$ .
Suppose that at time $t = 0,$ the position of mass is $x (0)$ and velocity $v (0),$ then its displacement can also be represented as $x (t) = C \cos (ω t - ϕ),$ where $C$ and $ϕ$ are

Question

The motion of a mass on a spring, with spring constant $K$ is as shown in figure.

The equation of motion is given by, $x (t) = A \sin ω t +$ $B \cos ω t$ with $ω = \sqrt{\frac{K}{m}}$ .
Suppose that at time $t = 0,$ the position of mass is $x (0)$ and velocity $v (0),$ then its displacement can also be represented as $x (t) = C \cos (ω t - ϕ),$ where $C$ and $ϕ$ are

Aicharya · Accepted Answer

x = A s i n &#969; t + B c o s &#969; t v = d x d t = A &#969; cos &#969; t - B &#969; sin &#969; t At t = 0 , &#160; x ( 0 ) = B v ( 0 ) = A &#969; x = A sin &#969; + B sin &#969; t + 90 &#176; A net = A 2 + B 2 tan &#945; = B A &#8658; cot &#945; = A B &#8658; x = A 2 + B 2 sin ( &#969; t + &#945; ) &#8658; x = A 2 + B 2 cos ( &#969; t - ( 90 - &#945; ) ) x = C cos ( &#969; t - &#981; ) &#8658; C = A 2 + B 2 C = [ v ( 0 ) ] 2 &#969; 2 + [ x ( 0 ) ] 2 &#981; = 90 - &#945; tan &#945; = cos &#945; = A B &#8658; tan &#981; = v ( 0 ) x ( 0 ) &#183; &#969; &#981; = tan - 1 v ( 0 ) x ( 0 ) &#969;

The motion of a mass on a spring, with spring constant K is as shown in figure. The equation of motion is…

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