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Chapter 15: Q.1 (page 414)

A block oscillating on a spring has period $T = 2 s$ . What is the period if:
a. The block's mass is doubled $?$ Explain. Note that you do not know the value of either $m$ or $k$ , so do not assume any particular values for them. The required analysis involves thinking about ratios.
b. The value of the spring constant is quadrupled?
c. The oscillation amplitude is doubled while and $k$ are unchanged?

Short Answer

Expert verified

a.The period is $2.828 s$ when block's mass is doubled.

b.The time if the spring constant is quadrupled is $1 s$ .

c.The oscillation amplitude is double while $m$ and $k$ change time period.

Step by step solution

01

Formula for time

Use the following Formula to find the time.

$T = 2 蟺 \sqrt{\frac{m}{k}}$

Here, $m$ is the mass and $k$ is the spring constant.

02

Calculation of time (part a)

The initial time

$T = 2 蟺 \sqrt{\frac{m}{k}} 鈰� 鈰� 路 (1)$

If the block mass is blocked, then $m = 2 m$ and new time $T^{'}$

$T^{'} = 2 蟺 \sqrt{\frac{2 m}{k}}$

$T^{'} = \sqrt{2} [2 蟺 \sqrt{\frac{m}{k}}]$

From equation $(1)$ the above equation will be,

$T^{'} = \sqrt{2} T$

Substitute $2 s$ for $T$ in the above equation

$T^{'} = \sqrt{2} (2 s)$

$T^{'} = 2.828 s$

03

Calculation for time if the spring constant is quadrupled (part b)

b)

If the value of spring constant is quadrupled, then new time $T^{'}$ is,

$T^{'} = 2 蟺 \sqrt{\frac{m}{4 k}}$

$T^{'} = \frac{1}{2} [2 蟺 \sqrt{\frac{m}{k}}]$

$T^{'} = \frac{1}{2} T$

Substitute $2 s$ for $T$ in the above equation.

$T^{'} = \frac{1}{2} (2 s)$

$T^{'} = 1 s$

04

Oscillation amplitude changes (part c)

c) Time doesn't depend on oscillation amplitude, until and unless $m$ and $k$ change time period.

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