
The period of a spring-mass system is directly proportional to the square root of the mass. Originally, the period of the system was $2.0$ seconds. The original mass was $1.0$ Kg.
Which value shown is closest to the new period if the new mass is $24.0$ Kg.
a. $48$ seconds
b. $24$ seconds
c. $10$ seconds
d. $22$ seconds
e. $8$ seconds
Answer
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Hint: As we are given, that the time period of a spring-mass system is directly proportional to the square root of the mass of the system, so, if we know the original mass and time period of the system and are given the new mass, we can easily calculate the new time period of the system for the new mass.
Complete step by step answer:
Here in the problem, it is given that, ${M_1} = 1.0$ Kg,
${T_1} = 2.0$ seconds,
And, ${M_2} = 24.0$ Kg,
Where, ${M_1}$ = original mass of the spring-mass system,
${M_2}$ = new mass of the spring-mass system,
And, ${T_1}$ = original time-period of the spring-mass system.
Now, let ${T_2}$ be the time-period of the spring-mass system when its mass is $24.0$ Kg.
We know, the time-period of a spring-mass system is directly proportional to the square root of the mass of the system, i.e., $T \propto \sqrt M $ .
So, we can say, $\dfrac{{{T_2}}}{{{T_1}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} $ .
Now, using the given values, we get $\dfrac{{{T_2}}}{{2.0}} = \sqrt {\dfrac{{24.0}}{{1.0}}} $ .
Therefore, ${T_2} = 9.8$ .
Now, we can see, this value of the new time period, if the mass is $24.0$ Kg, is closest to $10$ seconds from the given options.
Hence, the correct answer is option (C).
Additional information:
If the frequency of a spring-mass system be $f$ , and the spring constant be $k$ , then we know that $\dfrac{k}{m} = {(2\pi f)^2}$ .
Therefore, $f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} $ .
So, the time period of the system is, $T = \dfrac{1}{f}$ .
Then, $T = 2\pi \sqrt {\dfrac{m}{k}} $ .
Note: When a system of masses is connected using springs with several degrees of freedom, the total system is called a spring-mass system. If a system consists of two masses and three springs, then we can say the degrees of freedom for that system is $2$.
Complete step by step answer:
Here in the problem, it is given that, ${M_1} = 1.0$ Kg,
${T_1} = 2.0$ seconds,
And, ${M_2} = 24.0$ Kg,
Where, ${M_1}$ = original mass of the spring-mass system,
${M_2}$ = new mass of the spring-mass system,
And, ${T_1}$ = original time-period of the spring-mass system.
Now, let ${T_2}$ be the time-period of the spring-mass system when its mass is $24.0$ Kg.
We know, the time-period of a spring-mass system is directly proportional to the square root of the mass of the system, i.e., $T \propto \sqrt M $ .
So, we can say, $\dfrac{{{T_2}}}{{{T_1}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} $ .
Now, using the given values, we get $\dfrac{{{T_2}}}{{2.0}} = \sqrt {\dfrac{{24.0}}{{1.0}}} $ .
Therefore, ${T_2} = 9.8$ .
Now, we can see, this value of the new time period, if the mass is $24.0$ Kg, is closest to $10$ seconds from the given options.
Hence, the correct answer is option (C).
Additional information:
If the frequency of a spring-mass system be $f$ , and the spring constant be $k$ , then we know that $\dfrac{k}{m} = {(2\pi f)^2}$ .
Therefore, $f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} $ .
So, the time period of the system is, $T = \dfrac{1}{f}$ .
Then, $T = 2\pi \sqrt {\dfrac{m}{k}} $ .
Note: When a system of masses is connected using springs with several degrees of freedom, the total system is called a spring-mass system. If a system consists of two masses and three springs, then we can say the degrees of freedom for that system is $2$.
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