
Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass=500g, decay constant=20g/s then how much time is required for the amplitude of the system is to drop to half of its initial value?
(ln2=0.693)
a. 34.65 s
b. 17.32s
c. 0.034s
d. 15.01s
Answer
163.5k+ views
Hint:In simple harmonic motion, the amplitude remains constant most of the time. But there are situations in which some kind damping force is present which may be a complicated function of speed. By solving the equation for damped oscillation we get the equation connecting distance of particle at a time t and initial distance solving which we get the required time for the amplitude of the system to drop to half its initial value.
Formula used:
we have
$A={{A}_{0}}{{e}^{\frac{-\lambda t}{2m}}}$
where $\lambda$=decay constant or damping constant and ‘t’ is the time required and m is the mass of the substance.
Complete answer:
Let ‘A’ be the amplitude of the oscillation at a time ‘t’ and A0 be the initial amplitude. Then we have the equation,$A={{A}_{0}}{{e}^{\frac{-\lambda t}{2m}}}$ where $\lambda $=decay constant or damping constant and ‘t’ is the time required and m be the mass of the substance.
We have for damped simple harmonic motion, the amplitude at a time t is
$A={{A}_{0}}{{e}^{\frac{-\lambda t}{2m}}}$
Taking natural logarithm on both sides,
$\frac{\lambda t}{2m}=-\ln 2$
By substituting the value of ln2=0.693 we get:
Time taken for the system to drop to half of its initial value is
$t=\dfrac {-0.693\times 2\times 500}{20}$
On calculating we get required time, $t=34.6s$
Therefore, the Answer is (a)
Note: When you take the logarithm remember that you have to take natural logarithm and by putting value of \[A=\frac{{{A}_{O}}}{2}\] we can find the time required easily.
In simple harmonic motion, force is directly proportional to the distance and its direction is opposite to that of distance. But there are situations in which some kind of damping force is present which may be a complicated function of speed. Here damping force is directly proportional to velocity and directed opposite to velocity. Energy is lost due to the negative work and amplitude gradually decreases with time and finally reaches zero.
Formula used:
we have
$A={{A}_{0}}{{e}^{\frac{-\lambda t}{2m}}}$
where $\lambda$=decay constant or damping constant and ‘t’ is the time required and m is the mass of the substance.
Complete answer:
Let ‘A’ be the amplitude of the oscillation at a time ‘t’ and A0 be the initial amplitude. Then we have the equation,$A={{A}_{0}}{{e}^{\frac{-\lambda t}{2m}}}$ where $\lambda $=decay constant or damping constant and ‘t’ is the time required and m be the mass of the substance.
We have for damped simple harmonic motion, the amplitude at a time t is
$A={{A}_{0}}{{e}^{\frac{-\lambda t}{2m}}}$
Taking natural logarithm on both sides,
$\frac{\lambda t}{2m}=-\ln 2$
By substituting the value of ln2=0.693 we get:
Time taken for the system to drop to half of its initial value is
$t=\dfrac {-0.693\times 2\times 500}{20}$
On calculating we get required time, $t=34.6s$
Therefore, the Answer is (a)
Note: When you take the logarithm remember that you have to take natural logarithm and by putting value of \[A=\frac{{{A}_{O}}}{2}\] we can find the time required easily.
In simple harmonic motion, force is directly proportional to the distance and its direction is opposite to that of distance. But there are situations in which some kind of damping force is present which may be a complicated function of speed. Here damping force is directly proportional to velocity and directed opposite to velocity. Energy is lost due to the negative work and amplitude gradually decreases with time and finally reaches zero.
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