Question

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is 15 cm/sec and the period is 628 milli-seconds. The amplitude of the motion in centimetres is
A. 3.0
B. 2.0
C. 1.5
D. 1

Answer 1

Hint: Maximum speed or velocity is given by product of amplitude and angular acceleration, at maximum displacement velocity is zero and the displacement is called Amplitude.

Formula used :
Maximum speed or velocity,
\[{v_{max}} = A\omega \,\,\]
Angular frequency,
\[\omega \,\, = \,\dfrac{{2\Pi }}{T}\]
Here, A = Amplitude and T = Time period

Complete step by step solution:
Given here,
Maximum speed, \[{v_{max}} = 15\,cm/\sec \,\,\] Time period, T = 628 milli-seconds
Amplitude A =?
Relation between maximum velocity and amplitude is given by equation,
\[{v_{max}} = A\omega \,\,\]
Where, $\omega$ is the angular frequency and it is mathematically given by,
\[\omega \,\, = \,\dfrac{{2\Pi }}{T}\]
Where T is the time period.

Therefore maximum velocity can be expressed as,
\[{v_{max}} = A\dfrac{{2\Pi }}{T}\] and amplitude \[A = \dfrac{{T{v_{max}}}}{{2\Pi }}\]
Substituting \[{v_{max}} = 15\,cm/\sec \,\,\] and \[T = 628\,{\rm{milli - seconds}}\, = 628\, \times {10^{ - 3}}\,\sec \,\] in amplitude equation we get,
\[A = \dfrac{{628 \times {{10}^{ - 3}} \times 15}}{{2 \times 3.14}} \\
\therefore A = 1.5\,cm\]
Hence, the amplitude of motion is 1.5 cm.

Therefore, option C is the correct answer.

Note: From the solution it is clear that amplitude is directly proportional to the time period T and maximum velocity $v_{max}$, therefore any change in both of these quantities will directly affect the amplitude.